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					Chương II: Nguyên hàm và tích phân − Tr n Phương


             BÀI 8. PHƯƠNG PHÁP TÍCH PHÂN T NG PH N

I. CÔNG TH C TÍCH PHÂN T NG PH N

Gi s       u = u ( x ) ; v = v(x) có       o hàm liên t c trong mi n D, khi ó ta có:

                             ∫             ∫           ∫
• d ( uv ) = udv + vdu ⇔ d ( uv ) = udv + vdu ⇔ uv = udv + vdu         ∫       ∫
                                       b                        b
                                                           b
      ⇒    ∫ udv = uv − ∫ vdu ⇒ ∫ udv = ( uv )
                                       a
                                                           a    ∫
                                                               − vdu
                                                                a

Nh n d ng: Hàm s dư i d u tích phân thư ng là tích 2 lo i hàm s khác nhau
Ý nghĩa:         ưa 1 tích phân ph c t p v tích phân ơn gi n hơn (trong nhi u
trư ng h p vi c s d ng tích phân t ng ph n s kh b t hàm s dư i d u tích
phân và cu i cùng ch còn l i 1 lo i hàm s dư i d u tích phân)

Chú ý: C n ph i ch n u, dv sao cho du ơn gi n và d tính ư c v                                    ng th i

           tích phân    ∫ vd u     ơn gi n hơn tích phân             ∫ udv
II. CÁC D NG TÍCH PHÂN T NG PH N CƠ B N VÀ CÁCH CH N u, dv
1. D ng 1:
                                 u = P ( x )
            sin ( ax + b ) dx 
            cos ( ax + b ) dx          sin ( ax + b ) dx
                                        cos ( ax + b ) dx
 ∫   P (x) 
                    e
                       ax + b   ⇒
                              dx  dv =            ax + b
                                                                    (trong ó P(x) là a th c)
                      ax + b                    e        dx
                   m         dx                  ax + b
                                               m         dx
2. D ng 2:
                                     dv = P ( x ) dx
            arcsin ( ax + b ) dx 
            arccos ( ax + b ) dx        arcsin ( ax + b )
                                         arccos ( ax + b )
                arctg ( ax + b ) dx     
 ∫   P (x)                        ⇒        arctg ( ax + b )
                      ( ax + b ) dx u = 
                                                               (trong ó P(x) là a th c)
            arc cotg                
                                         arc cotg ( ax + b )
                   ln ( ax + b ) dx      ln ( ax + b )
           
            log m ( ax + b ) dx 
           
                                     
                                          
                                          log m ( ax + b )
                                         
3. D ng 3:
                           sin ( ln x )              eax+b sin ( αx + β) dx  eax+b
      sin ( ln x ) dx     cos ( ln x )
       ( )                                            ax+b                   u =  ax+b
     k cos ln x dx        u = sin ( log x )          e cos ( αx + β) dx ⇒  m
∫   x 
      
                        ⇒
       sin ( loga x) dx  cos ( log x)
                          
                                         a    ;   ∫    max+b sin ( αx + β) dx 
                                                                               dv = sin ( αx + β) dx
      cos ( loga x ) dx   k                         ax+b                         cos ( αx + β) dx
                                         a
      
                          dv = x dx                  m cos ( αx + β) dx 
                                                                                   


210
                                                                         Bài 8. Phương pháp tích phân t ng ph n


III. CÁC BÀI T P M U MINH H A:



                ∫ P ( x ) {sin ( ax + b ) ; cos ( ax + b ) ; e                      ; m ax+b } dx
                                                                             ax+b
1. D ng 1:



         ∫
• A1 = x 3 cos x dx .

                                       u = x 3
                                                     du = 3x 2 dx
                                                      
Cách làm ch m:                       t              ⇒             . Khi ó ta có:
                                       dv = cos x dx  v = sin x
                                                     
                                                        u = x 2
                                                                      
                                                                       du = 2x dx
                        ∫
           3                    2
A1 = x sin x − 3 x sin x dx .                         t              ⇒           . Khi ó ta có:
                                                        dv = sin x dx  v = −cosx
                                                                      
                                                                               u = x
                                                                                             du = dx
                                                                                              
A1 = x sin x − 3  − x cos x + 2 x cos x dx  .   ∫
      3               2
                                                                             t              ⇒           .
                                                                             dv = cos x dx  v = sin x
                                                                                             
       3            2
                                          (           ∫      )           3                2
A1 = x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x ) + c

Cách làm nhanh: Bi n                          i v d ng     ∫ P ( x ) L ( x ) dx = ∫ P ( x ) du
A1 = x3 cos x dx = x 3 d ( sin x ) = x 3 sin x − sin x d ( x3 ) = x3 sin x − 3 x 2 sin x dx
      ∫                         ∫                                    ∫                                  ∫
    = x sin x + 3 x d ( cos x ) = x sin x + 3  x cos x − cos x d ( x ) 
                     ∫                                                                ∫
       3           2               3             2                   2
                                                                       

                                              ∫                                                     ∫
          3                 2                                    3                   2
    = x sin x + 3x cos x − 6 x cos x dx = x sin x + 3x cos x − 6 x d ( sin x )
         3          2
                                          (           ∫      )           3                2
    = x sin x + 3x cos x − 6 xsin x − sin xdx = x sin x + 3x cos x − 6 ( xsin x + cos x) + c

                                    1 3 ( 5 x −1 ) 1  3 5 x −1
          ∫
• A2 = x 3 e 5x − 1 dx =              ∫
                                      x d e       = xe          − e5 x −1 d ( x3 ) 
                                                                              ∫
                                    5              5                              
   1                                  1              3 2 ( 5x −1 ) 
  =  x 3 e5x −1 − 3 x 2 e5x −1 dx  =  x 3 e5x −1 −
                        ∫                               x d e           ∫
   5                               5               5             
   1            3                                   1            3             6
  = x 3 e5x −1 −  x 2 e5x −1 − e5x −1 d ( x 2 )  = x 3 e5x −1 − x 2 e5x −1 +
                                              ∫                                   xe5x −1 dx                ∫
   5            25                               5             25            25
   1            3              6                  1            3
  = x 3 e5x −1 − x 2 e5x −1 +
   5            25            125                 ∫
                                  x d ( e5x −1 ) = x 3 e5x −1 − x 2 e5x −1 +
                                                  5            25
           6  5x −1                1            3              6             6 5x −1
     +          xe   − e5x −1 dx  = x 3 e5x −1 − x 2 e5x −1 +
                                ∫                                  xe5x −1 −     e    +c
          125                    5             25            125           625
Nh n xét: N u P(x) có b c n thì ph i n l n s d ng tích phân t ng ph n.


                                                                                                                211
Chương II: Nguyên hàm và tích phân − Tr n Phương

          π2 /4                                                                                                         x           0            π2/4
• A3 =        ∫       x sin x dx .                                 t t = x ⇒t = x ⇒        2
                                                                                                                        t           0         π/2
              0
                                                                                                                        dx                  2tdt
          π2                                   π2                                                          π2                               π2
                                                                                                   π2
          ∫                                        ∫                                                    + 2 ∫ cos td ( t ) = 6 ∫ t cos t dt
                  3                                        3                           3                                3         2
A3 = 2 t sin t dt = −2 t d ( cos t ) = −2t cos t 0
          0                                        0                                                        0                               0

   π2                                                               π2                                      π2                                     π2
                                                       π2                                      3π2                  3π2
= 6 t 2 d ( sin t ) = 6t 2 sin t 0 − 6 sin td ( t 2 ) =
      ∫                                                              ∫                                          ∫
                                                                                                   −12 t sin t dt =     + 12 td ( cos t )           ∫
      0                                                              0
                                                                                                2     0
                                                                                                                     2      0

                                                           π2
  3π 2             π2                                                              3π2            π2  3π2
=
   2
       + 12t cos t 0 − 12                                      ∫
                                                               0
                                                                   cos t dt =
                                                                                    2
                                                                                       − 12 sin t 0 =
                                                                                                       2
                                                                                                          − 12


          π6                                                             π 6                                                π 6           π 6
                                      1                                                           x cos3 x                            1
          ∫                                                              ∫     x d ( cos3 x ) = −                                           ∫ cos
                                           2
• A4 =            x sin x cos xdx = −                                                                                               +               3
                                                                                                                                                        x dx
          0
                                      3                                   0
                                                                                                                    3       0         3     0


                              π6                                                                                                            π6
                                                                                        
                                   (1 − sin x ) d ( sin x ) = − π 3 + 1  sin x − sin x  = 11π − π 3
                                                                                      3
    π 3 1
                              ∫
                                                       2
 =−    +                                                                
     48 3                     0
                                                                 48 3               3 0    72    48

          1
                                                     u = x 2 e x     du = x ( x + 2 ) e x dx
                      2   x
               x e dx                                                
• A5 =    ∫ ( x + 2)
          0
                               2
                                   .               t 
                                                     dv =
                                                                 dx ⇒ 
                                                                      v = −
                                                                                1
                                                          ( x + 2 )2       x+2

                      2 x 1            1                                       1                            1
        x e            e           e
            + xe dx = − + xe dx = − + xd ( e )
                                       ∫                                       ∫                            ∫
                x           x               x
 A5 = −
        x+20 0         3 0         3 0
                                   1           1                                               1
         e                                       e                                                          e
                                           ∫
               x                          x               x
      = − + xe                     0
                                       − e dx = − + e − e                                      0
                                                                                                   =1−
         3                              0
                                                 3                                                          3

2. D ng 2:
                          ∫ P ( x ) {arcsin u; arccos u; arctg u; arc cotg u ; ln u ; log                                               m   u u = ax + b} dx
          e
                                                       1
                                                         e
                                                                                                  e  e
                                                                                                                       
                                   2
                                                           ( ln x ) 2 d ( x 3 ) = 1  x3 ( ln x )2 1 − x3 d ( ln x ) 2 
          ∫
• B1 = x 2 ( ln x ) dx =
          1
                                                       31  ∫                      3                 1                
                                                                                                                       
                                                                                                                                ∫
       1                dx  1  3               1 3 2 e                  
              e                      e
      =  e3 − 2x 3 ln x  ∫
                                        2
                             =  e − 2x ln x dx  =  e −    ln x d ( x 3 )      ∫                                                 ∫
       3
             1
                          x  3
                                   1
                                                  3
                                                         31                
                                                                             

     e 2 ( 3                       3
      3            e  e                            e         3      e    3
    = −  x ln x ) 1 − x3d ( ln x ) = e − 2 e3 + 2 x2 dx = e + 2 x3 = 5e − 2
                                                       ∫                                                   ∫
     3 9            1             3 9
                                                 91        9 27 1 27 27

212
                                                                                       Bài 8. Phương pháp tích phân t ng ph n


                                                           1 + x  ( 2 ) 1  2 1 + x             1 + x 
         12                                    12                                        12 12
                    1+ x       1                                                             2 
• B2 =   ∫
         0
              x ln        dx =
                    1− x       2             ∫0
                                                       ln 
                                                           1− x 
                                                                   d x =  x ln
                                                                          2 
                                                                            
                                                                                        − x d  ln
                                                                                  1− x  0   0
                                                                                                          
                                                                                                  1 − x 
                                                                                                                                        ∫
                      12                                                          12                   2
      1                             1 − x dx     1                                      x 
                      ∫                                                           ∫
                            2
     = ln 3 −              x ⋅           ⋅      = ln 3 −                                     dx
      8               0
                                    1+ x 1− x 2
                                                 8                                0
                                                                                       1+ x 
                      12                       2                             12
      1                        1      1                                                         1                    2 
     = ln 3 −
      8               ∫
                      0
                           1 −    dx = ln 3 −
                            1+ x      8                                    ∫ 1 + (1 + x )
                                                                             0 
                                                                                                           2
                                                                                                                   −
                                                                                                                       1+ x 
                                                                                                                            
                                                                                                                              dx

                                                                        12
      1             1                   ln 3       3 5
     = ln 3 −  x −      − 2 ln 1 + x  =      + 2 ln −
      8            1+ x              0   8         2 6
         1                                                                                     1       1
• B3 = ∫ ln ( x +          1+ x         2   ) dx =  x ln ( x +
                                                                            1+ x      2   ) 0 − ∫ xd ln ( x +
                                                                                                                                              1 + x2 )
                                                                                                                                                        
         0                                                                                             0

                                1                                                                                                   1
                             x        dx                       x dx
     = ln (1 + 2 ) − x  1 +    ∫              = ln (1 + 2 ) −                                                                      ∫
                                 2          2                        2
                    0       1+ x  x + 1+ x                   0 1+ x


                                1 d (1 + x 2 )
                                    1                                                                          1
     = ln (1 + 2 ) −                ∫          = ln (1 + 2 ) − 1 + x                                               = ln (1 + 2 ) + 2 − 1
                                                                     2
                                                                                                               0
                                2 0 1+ x   2



              x ln ( x + 1 + x 2              ) dx = 1 ln ( x +
         1
• B4 =   ∫                                           ∫                           1 + x2 ) d ( 1 + x2                        )
                                    2
         0             1+ x                          0

                                                           1       1
                      ln ( x +                         )           ∫    1 + x d  ln x + 1 + x
                                                                                           (                                 
                                                                                                                            )
                  2                                2                          2                                         2
     = 1+ x                             1+ x                   −                
                                                           0
                                                                   0
                                                                                                                               
                                        1
                                                                    dx
     = 2 ln (1 + 2 ) −                  ∫     1 + x 1 +  x    
                                                   2
                                                            2          2
                                        0               1+ x  x + 1+ x
                                        1
     = 2 ln (1 + 2 ) − dx = 2 ln (1 + 2 ) − 1
                                        ∫
                                        0


                                                                     u = ln x + 1 + x 2
                                                                                   (                               )
              x ln ( x + 1 + x 2              ) dx .
         1
                                                                     
• B5 =   ∫
         0       x + 1 + x2
                                                                   t 
                                                                     dv =
                                                                               x dx    =x                              (        1 + x − x dx
                                                                                                                                            2
                                                                                                                                                    )
                                                                                     2
                                                                           x + 1+ x

               x                                     (x +                  )             dx
  ⇒ du =  1 +                           dx                           1 + x2 =
                    2                                                                              2
              1+ x                                                                    1+ x


                                                                                                                                                            213
Chương II: Nguyên hàm và tích phân − Tr n Phương


  v=
          1
                  ∫ (1 + x 2 )1 2 d (1 + x 2 ) − ∫ x 2 dx = 1 (1 + x 2 )3 2 − x 3 
                                                                                  
          2                                                                                         3

                                                                                                            1             1
     1                            2 
B5 =  (1 + x ) − x  ln x + 1 + x  −
              2 32                      1 (      2 32  3  dx      (                               )
                                             1+ x ) − x                                                                 ∫
                    3
                     
     3                              0 3 0                1+ x
                                                                2



          (2          2 − 1) ln (1 + 2 ) 1      dx    1 x 3 dx
                                                                        1                               1
      =
                            3
                                        −           +
                                          3 0 1 + x2 3 0 1 + x2         ∫                               ∫
          (2          2 − 1) ln (1 + 2 ) 1          1 (1 + x ) − 1 (
                                                                                            1           1                     2
                                                                  d 1+ x )                              ∫
                                                                        2
      =                                 − arctg x +
                            3            3       0  60    1+ x 2



          (2          2 − 1) ln (1 + 2 ) π 1 (         2 12       2 −1 2 
                                                                                        1

                                                  1 + x ) − (1 + x )  d (1 + x )      ∫
                                                                                2
      =                                 −   +
                            3             12 6 0

          (2          2 − 1) ln (1 + 2 ) π 1  2 (   2 32         2 12
                                                                                                                                                              1

      =                                 −   +  1 + x ) − 2 (1 + x ) 
                            3             12 6  3                    0
          (2          2 − 1) ln (1 + 2 ) π 2 − 2
      =                                 −    +
                            3             12   9
              1                                                                     1
• B6 = ∫ x ln ( x +                           1+ x      2   ) dx = 1 ∫ ln ( x +                             1 + x2 ) d ( x2 )
              0
                                                                                2   0

                                                             1
        x 2 ln x + 1 + x 2   (                        )       1 2 
                                                                                1
      =                                                          x d  ln x + 1 + x 
                                                                                ∫                   (                                  )
                                                                                     2
                                                              −                        
       
                 2                                         0 2
                                                                0

                                                   1                                                                                                                       1    2
          1               1 2    x        dx      1              1                                                                                                           x dx
      =     ln (1 + 2 ) −   x 1+                   = ln (1 + 2 ) −
                                                   ∫                                                                                                                       ∫
          2               20 
                                     2 
                                 1+ x  x + 1+ x
                                                 2  2              2                                                                                                       0   1+ x
                                                                                                                                                                                      2



                      1                                                                                                           x              0                 1
                                  x 2 dx
 Xét I =              ∫
                      0           1 + x2
                                              .        t x = tg t ; t ∈  0, π ⇒
                                                                         2
                                                                                                       )                         t              0             π/4
                                                                                                                                  dx             dt cos 2 t
          1               2                   π4            2                                       π4                2                    π4        2
                      x dx                             tg t                         dt                      sin t                               sin t
⇒I=       ∫
          0           1+ x
                                      2
                                          =   ∫
                                              0    1 + tg t cos t
                                                                    2
                                                                            ⋅           2
                                                                                                =   ∫ cos
                                                                                                    0
                                                                                                                      3
                                                                                                                          t
                                                                                                                              dt =         ∫ cos
                                                                                                                                           0
                                                                                                                                                     4
                                                                                                                                                         t
                                                                                                                                                             d ( sin t )

              π4                  2                             2 2             2                               2 2                                            2
                      sin t d ( sin t )                                     u du                  1                    (1 + u ) − (1 − u ) 
          =       ∫
                  0       (1 − sin 2 t )2
                                                       =        ∫
                                                                0       (1 − u 2 )2
                                                                                                =
                                                                                                  4             ∫
                                                                                                                0
                                                                                                                       (1 + u ) (1 − u )  du
                                                                                                                                           


214
                                                                                                       Bài 8. Phương pháp tích phân t ng ph n

               2 2                                         2                 2 2
       1              1    1         1                                                1              1       2 
     =
       4       ∫0
                         −      du =
                     1− u 1+ u       4                                     ∫   0
                                                                                       (
                                                                                        1 − u)
                                                                                                2
                                                                                                  +           −
                                                                                                    (1 + u ) 1 − u 2 
                                                                                                            2
                                                                                                                     
                                                                                                                       du

                                                                                 2 2
      1 1     1          1+ u                                                                     2
     =      −     − 2 ln                                                             =              − ln (1 + 2 )
      4 1− u 1+ u        1− u  0                                                                 2

      1             1   1             1 2                     2
⇒ B6 = ln (1 + 2 ) − I = ln (1 + 2 ) −    − ln (1 + 2 )  = −    + ln (1 + 2 )
      2             2   2             2 2                    4
         0                                            0                                                                                0
                                                    1                       1            0    1
• B7 =   ∫    x ln 1 − xdx =                          ∫  ln 1 − x d ( x2 ) = x2 ln 1 − x −8 −      x 2 d ( ln 1 − x )                  ∫
         −8
                                                    2 −8                    2                 2 −8
                                0                                                                                       0   2
                           1           −1    dx               1 x dx
    = −32ln 3 −
                           2 −8 ∫
                                x2 ⋅       ⋅
                                     2 1− x 1− x
                                                 = −32 ln 3 +
                                                              4 −8 1 − x                                                ∫
                           1 1 − (1 − x                            )
                                0                              2                                               0
                                                                                                        1  1                  
    = −32ln 3 +                 ∫
                           4 −8 1 − x
                                                                       dx = −32 ln 3 +                         ∫
                                                                                                        4 −8 1 − x
                                                                                                             
                                                                                                                    − (1 + x )  dx
                                                                                                                               
                                                                                       0
                           1                  1 2                   l            63
    = −32ln 3 +              − ln 1 − x − x − 2 x  = −32 ln 3 + 6 + 2 ln 3 = 6 − 2 ln 3
                           4                       −8

         0                                                                                                         x        −3             0
                    ln 1 − x
• B8 =   ∫ (1 − x )
         −3                     1− x
                                                    dx .               t t = 1− x ⇒                                 t           2          1
                                                                                                                   dx               −2tdt
                                    1                                        2                             2
Khi ó ta có: B8 =                   ∫
                                    2
                                        ln t
                                            t
                                                3
                                                     ( −2t dt ) = 2 ln t dt = 2 ln t d −1
                                                                   ∫ 2 ∫     1             t               1
                                                                                                                        (t)
                       2                2                                                          2                            2
           −2 ln t      −1                          dt          2
         =
             t 1
                   −2
                      1
                         t          ∫
                           d ( ln t ) = − ln 2 + 2 2 = − ln 2 −
                                                  1 t
                                                                t1
                                                                   = 1 − ln 2                  ∫
         3
                                  1 ln x d ( x 2 + 1) 1
                                                3                                              3
              x ln x dx                                           −1 
• B9 =   ∫ (x
         1
                2
                     + 1)
                            2
                                =
                                  2 1 ( x 2 + 1)∫2
                                                     =
                                                       21
                                                          ln x d  2    
                                                                  x + 1                  ∫
                           3                3                                                                  3
              − ln x              1     1               − ln 3 1         dx
    =
         2 ( x + 1) 1
                2
                                +      2
                                  2 1 x +1  ∫
                                           d ( ln x ) =
                                                         20
                                                              +
                                                                2 1 x ( x 2 + 1)                               ∫
         − ln 3 1 ( x + 1) − x
                           3        2                          2                                       3
                                     − ln 3 1  1      x 
    =
          20
               +           ∫
                 2 1 x ( x + 1)
                          2
                                dx =
                                      20
                                           +      − 2    dx
                                             2 1  x x +1                                             ∫
                                                                         3
      − ln 3 1      1 2          9 ln 3
    =       + ln x − ( x + 1)  =        −2
       20    2      2         1   20


                                                                                                                                               215
Chương II: Nguyên hàm và tích phân − Tr n Phương

3. D ng 3: Tích phân t ng ph n luân h i

                                                1                        1                   1 3
            ∫
• C 1 = x 2 sin ( ln x ) dx =
                                                3   ∫
                                                  sin ( ln x ) d ( x3 ) = x 3 sin ( ln x ) −
                                                                         3                   3
                                                                                               x d ( sin ( ln x ) )        ∫
 1                 1               dx 1                1
                   3                 ∫
= x3 sin ( ln x ) − x3 cos ( ln x ) = x3 sin ( ln x ) − x2 cos ( ln x ) dx
 3                                  x 3                3                                 ∫
 1 3              1                      1 3              1 3              1 3
= x sin ( ln x ) − cos ( ln x ) d ( x ) = x sin ( ln x ) − x cos ( ln x ) + x d ( cos ( ln x ) )
                                     ∫                                                                                         ∫
                                     3

 3                9                      3                9                9
 1 3              1 3              1 2                1 3              1 3              1
                                                        ∫
= x sin ( ln x ) − x cos ( ln x ) − x sin ( ln x) dx = x sin ( ln x ) − x cos ( ln x ) − C1
 3                9                9                  3                9                9
    10     1                 1                       1
⇒      C1 = x3 sin ( ln x ) − x3 cos ( ln x ) ⇒ C1 = 3x3 sin ( ln x ) − x3 cos ( ln x )  + c
                                                                                        
     9     3                 9                      10
            π                              π                                    π                π
                    1 2x                    e2 x                                            1 2x              e2π −1 1
            ∫                              ∫                                                     ∫
                    2x       2
• C2 = e sin x dx =    e (1 − cos 2x ) dx =                                             −      e cos 2 x dx =       − J
      0
                    20                       4                                  0           20                   4   2
        π                                 π                                                  π                π
                                         1 2x               1 2x        1
        ∫                                 ∫                                sin 2x d ( e )                     ∫
                                                                                       2x
J = e 2 x cos 2 x dx =                      e d ( sin 2x ) = e sin 2x −
        0
                                         20                 2        0  20
            π                                  π                                                  π               π
                                          1 2x               1              1
            ∫
    = − e 2x sin 2x dx =                       ∫
                                             e d ( cos 2x ) = e 2x cos 2x −    cos 2x d ( e 2x )                  ∫
            0
                                          20                 2           0  20
            2π               π                          2π                               2π                                 2π
        e         −1                 e −1            e −1     e −1
                             ∫
                        2x
    =                − e cos 2x dx =      − J ⇒ 2J =      ⇒J=
                 2    0
                                       2               2        4

                    e2 π − 1 1   e2π − 1 e2 π − 1 e2 π − 1
⇒ C2 =                      − J=        −        =
                        4    2      4        8        8
            eπ                                                     eπ                                                              eπ
                                                             eπ
            ∫ cos ( ln x ) dx =                                    ∫ xd ( cos ( ln x )) = − ( e                           + 1) +   ∫ sin ( ln x ) dx
                                                                                                                      π
• C3 =                                        x cos ( ln x ) 1 −
                1                                                  1                                                               1
                                 π                                                                            π
                             e                                                                            e
                                                                                              eπ
= − ( e + 1) + ∫ sin ( ln x ) dx = − ( e + 1) +                                                           ∫
                                                                                                      − xd ( sin ( ln x ) )
       π                                π
                                                                        x sin ( ln x )        1
                             1                                                                            1
                                 π
                             e
                                                                                                                                        −1 ( π )
= − ( e + 1) − cos ( ln x ) dx = − ( e + 1) − C3 ⇒ 2C3 = − ( e + 1) ⇒ C3 =
                             ∫
            π                                            π                                                π
                                                                                                                                            e +1
                             1
                                                                                                                                        2
            eπ                                 eπ                                   eπ               eπ
                          1                         1                                         1                    eπ −1 1
• C4 = ∫ cos ( ln x ) dx = ∫ [1 + cos ( 2ln x)] dx = x
                         2
                                                                                            −         ∫
                                                                                                 cos ( 2ln x) dx =      − I
            1
                                          2    1
                                                                            2       1         21                     2   2




216
                                                                                                      Bài 8. Phương pháp tích phân t ng ph n

                   eπ                                                                        eπ                                                    eπ
                                                                                    eπ                                                                  2sin ( 2lnx)
                    ∫
Xét I = cos ( 2 ln x ) dx = xcos( 2lnx) 1 − xd( cos( 2lnx) ) = e −1+ x                       ∫                                                     ∫
                                                                                                                                           π
                                                                                                                                                                     dx
                    1                                                                        1                                                     1
                                                                                                                                                              x
                                π                                                                                                      π
                            e                                                                                          π           e
                                                                                                                   e
= e − 1 + 2 ∫ sin ( 2 ln x ) dx = e − 1 + 2x sin ( 2 ln x ) 1 − 2 ∫ xd ( sin ( 2 ln x ) )
   π                               π

                            1                                                                                                      1
                                π                                                                 π
                            e                                                                 e
                                     2 cos ( 2 ln x )
                            ∫                                                                 ∫
   π                                                        π                             π
= e −1− 2 x                                           dx = e − 1 − 4 cos ( 2 ln x ) dx = e − 1 − 4I
                            1
                                           x                        1

                                                 eπ − 1                                eπ − 1 6 ( π
⇒ 5I = eπ − 1 ⇒ I =                                     ⇒ C4 = e π − 1 + I = e π − 1 +       = e − 1)
                                                   5                                     5    5
                            1 + sin x
• C5 = e x     ∫            1 + cos x
                                      dx =
                                           1 + sin x ( x ) x 1 + sin x
                                           1 + cos x
                                                     d e =e      ∫
                                                             1 + cos x
                                                                       − e x d 1 + sin x
                                                                               1 + cos x                                           ∫           (                  )
                                                                                                                                           x                 x
           1 + sin x     x 1 + cos x + sin x        x 1 + sin x     e dx       e sin x dx
                                         ∫                                                                                     ∫                        ∫
      x
 =e                  − e                  2
                                             dx = e             −           −
           1 + cos x         (1 + cos x )             1 + cos x   1 + cos x   (1 + cos x )2
                                                                                   x                               x
           1 + sin x                     e dx         e sin x dx
                                                                           ∫                               ∫
      x
 =e                  − I − J (1) ; I =           ;J=
           1 + cos x                   1 + cos x     (1 + cos x ) 2
                                                                       u = e x              du = e x dx
                            e x sin x dx                                                    
Xét J =             ∫ (1 + cos x )                   .               t         sin x dx ⇒         −d (1 + cos x )       1
                                                                                                                               ∫
                                                 2
                                                                       dv =               2 v =                   =
                                                                             (1 + cos x )          (1 + cos x ) 2
                                                                                                                      1 + cos x
                            x                        x                         x
                  e          e dx       e
⇒ J=
               1 + cos x
                         −          =        ∫
                           1 + cos x 1 + cos x
                                               −I                                                     (2). Thay (2) vào (1) ta có:

                            1 + sin x     ex                    x 1 + sin x    e
                                                                                   x
⇒ C5 = e x                            −I−           − I + c = e             −         +c
                            1 + cos x     1 + cos x               1 + cos x 1 + cos x
                                                     π                                                 π                           π
                   sin 2 x
               π
                                1 −x                     1 −x      1 −x
• C6 =         ∫
               0
                     e x
                           dx =
                                20                   ∫
                                   e (1 − cos 2 x ) dx =
                                                         20
                                                            e dx −
                                                                   20
                                                                      e cos 2 x dx                     ∫                           ∫
               −x π                  π                                             −π             π                                            −π
    −e                              1 −x              1− e                                   1 −x              1− e                                         1
  =
     2                  0
                            −        ∫
                                    20
                                       e cos 2 x dx =
                                                        2
                                                                                         −
                                                                                             20   ∫
                                                                                                e cos 2 x dx =
                                                                                                                 2
                                                                                                                                                        −
                                                                                                                                                            2
                                                                                                                                                              J

       π                                                 π                                   −x                π               π
                                 1 −x                e                                            sin 2x                   1
       ∫                                                 ∫                                                                    sin 2x d ( e )
                                                                                                                               ∫
               −x                                                                                                                         −x
J= e                cos 2 x dx =    e d ( sin 2x ) =                                                               −
       0
                                 20                                                                2           0           20
           π                                                 π                                                             π               π
   1 −x              −1 − x                  e− x cos 2 x     1
 =
   20      ∫
      e sin 2 x dx =
                     4 0
                         e d ( cos 2 x ) = −
                                                   4      0
                                                            +∫40
                                                                 cos 2 x d ( e − x )                                                       ∫


                                                                                                                                                                          217
Chương II: Nguyên hàm và tích phân − Tr n Phương

                             π
     1 − e −π 1 − x              1 − e −π 1   5   1 − e −π     1 − e −π
 =
        4
             −
               40            ∫
                  e cos 2 x dx =
                                    4
                                         − J ⇒ J=
                                          4   4      4
                                                           ⇒J=
                                                                  5

                  1 − e −π 1   1 − e −π 1 − e −π 2 (
⇒ C6 =                    − J=         −        = 1 − e −π )
                     2     2      2       10     5
          a
• C7 =    ∫
          0
                   a 2 − x 2 dx ; ( a > 0 )

                                                                                                                      a2 − ( a2 − x2 )
                                             a                                          a                         a
                                 a
                                                                                             x 2 dx
C7 = x a 2 − x 2                 0   − x d ( a2 − x2 ) =
                                             ∫                                         ∫                  =       ∫                      dx
                                             0                                          0    a2 − x2              0       a2 − x2
              a                                  a                                                    a       a
                        dx                                                                        x                                     πa 2
     = a2     ∫
              0        a2 − x2
                                         −       ∫
                                                 0
                                                         a 2 − x 2 dx = a 2 arcsin
                                                                                                  a   0
                                                                                                          −   ∫
                                                                                                              0
                                                                                                                       a 2 − x 2 dx =
                                                                                                                                         2
                                                                                                                                             − C7

                         2                                   2
                   πa        πa
⇒ 2C7 =               ⇒ C7 =
                    2         4
          a
• C8 =    ∫
          0
                   a 2 + x 2 dx ; ( a > 0 )

                                             a                                                    a
                                 a
                                          (                                      ) = a2                       x2
                                 0 − ∫ xd                                                         ∫
                  2          2                                       2       2
C8 = x a + x                                                     a +x                        2−                           dx
                                             0                                                    0       a2 + x2
                         a
                             ( a2 + x2 ) − a2                                                a                                 a
                                                                                                                                    dx
                         ∫                                                                   ∫                                 ∫
          2                                                                      2                2           2            2
     =a           2−                                                 dx = a            2−        a + x dx + a
                                             2           2
                         0           a +x                                                    0                                 0
                                                                                                                                    2
                                                                                                                                   a + x2
                                                                         a        a
     = a 2 2 + a 2 ln x + a 2 + x 2                                      0
                                                                             −   ∫      a 2 + x 2 dx = a 2 2 + a 2 ln (1 + 2 ) − C8
                                                                                  0


                                                                                            2 + ln (1 + 2 ) 2
⇒ 2C8 = a 2 2 + a 2 ln (1 + 2 ) ⇒ C8 =                                                                     a
                                                                                                  2
          a                                                                             u = x               du = dx
                                                                                                            
          ∫
                   2         2           2
• C9 = x                a + x dx ; ( a > 0 ) .                                        t                    ⇒                 3
                                                                                        dv = x a 2 + x 2 dx  v = 1 ( a + x ) 2
                                                                                                                        2   2
          0                                                                                                      3
               3                         a               a3
    x 2                                        1 ( 2
C9 = ( a + x ) 2                                  a + x ) 2 dx
                                                         ∫
            2                                          2
                                             −
    3                                    0     30
                                     a                                            a
         2 2 4 a2                                                            1 2 2           2 2 4 a2     1
                                     ∫                                            ∫
                                                     2           2                     2
     =      a −                                  a + x dx −                     x a + x dx =    a −   C8 − C9
          3     3                    0
                                                                             30               3     3     3



218
                                                                                           Bài 8. Phương pháp tích phân t ng ph n

                 2 + ln (1+ 2 ) 3 2 − ln (1+ 2 )        3 2 − ln (1+ 2 )
                                  2
 4      2 2 4 a
⇒ C13 =    a − ⋅               =                 ⇒ C9 =
 3       3     3       2                6                       8

            a                                                                  u = x               du = dx
                                                                                                   
            ∫
                  2     2     2
• C 10 = x             a − x dx ; ( a > 0 ) .                                t                    ⇒                  3
                                                                               dv = x a 2 − x 2 dx  v = −1 ( a − x ) 2
                                                                                                                2   2
            0                                                                                           3
                                  a
                                                             1                                     
                   3                       a           3        a                 a
      −x ( 2                              1 ( 2
C10 =     a − x2 ) 2                  +      a − x 2 ) 2 dx =  a 2 a 2 − x 2 dx + x 2 a 2 − x 2 dx 
                                                  ∫                                        ∫                                                ∫
       3                          0       30                 30
                                                                                 0
                                                                                                    
                                                                                                    
        2                                                        2                  4                                          4
       a      1                        2     a     πa                                                                  πa
   =      C7 + C10                    ⇒ C10 = C7 =                                             ⇒ C10 =
        3     3                        3      3    12                                                                   8
            2a                                                        2a            2a
• C 11 =     ∫        x 2 − a 2 dx = x x 2 − a 2                      a 2      −    ∫      x d ( x2 − a2 )
            a 2                                                                    a 2


                                                                                                                                            a + (x − a               )
                                          2a                                                                                           2a    2           2       2
                                                                x
       = (2 3 − 2 ) a −                   ∫                                  dx = ( 2 3 − 2 ) a −                                      ∫
                              2                                                                                            2
                                                  x                                                                                                                      dx
                                                            2         2                                                                              2       2
                                       a 2                 x −a                                                                    a 2           x −a
                                                  2a                               2a
                                                                dx
       = (2 3 − 2 ) a − a                         ∫                                ∫
                              2               2                                                2           2
                                                                              −            x − a dx
                                                             2           2
                                                  a 2       x −a                  a 2

                                                                                    2a             2a
       = ( 2 3 − 2 ) a − a ln x + x − a                                                            ∫
                              2               2                      2         2                                   2               2
                                                                                    a 2
                                                                                            −                  x − a dx
                                                                                                   a 2

                                                        2+ 3
       = ( 2 3 − 2 ) a − a ln
                              2               2
                                                                      − C11
                                                        1+ 2

                                                                                                       a                    2+ 3
                                                                                                         2
                                                                2+ 3
⇒ 2C11 = ( 2 3 − 2 ) a − a ln                                                                              ( 2 3 − 2 ) − ln
                                          2            2
                                                                                   ⇒ C11 =                                        
                                                             1+ 2                                       2                   1+ 2 
            π 2               π 2                                                                  π 2         π 2
                    dx
                                                                                                                ∫ cotg x d ( sin x )
                                1                    cotg x                                                                                      1
• C 12 =    ∫
            π 4
                     3
                  sin x
                        =−
                          π 4
                                  ∫
                              sin x
                                    d ( cotg x ) = −
                                                      sin x π                                          4
                                                                                                           +
                                                                                                               π       4
                       π2                                                          π2
                                          cos x                                            1
                                                                                   ∫ sin x  sin                            
                                                                                                               1
       =− 2 −           ∫ cotg x sin
                       π4
                                                  2
                                                      x
                                                           dx = − 2 −
                                                                                   π4
                                                                                           
                                                                                           
                                                                                                               2
                                                                                                                       x
                                                                                                                         − 1 dx
                                                                                                                            
                       π2                 π2                                        π2
                           dx         dx               sin x dx
       =− 2 +           ∫
                       π4
                               −
                          sin x π 4 sin 3 x   ∫
                                            =− 2 +
                                                   π4
                                                              2
                                                      1 − cos x
                                                                − C12                  ∫
                                                            π2
                            1 1 + cos x                                                                                                     − 2 + ln (1 + 2 )
⇒ 2C12 = − 2 −               ln                                      = − 2 + ln (1 + 2 ) ⇒ C12 =
                            2 1 − cos x                     π4                                                                                     2


                                                                                                                                                                              219
Chương II: Nguyên hàm và tích phân − Tr n Phương

4. D ng 4: Các bài toán t ng h p

              3
                                                                            x3 ( x 2 + 2 )                         x 3 ( x 2 + 1)
                                                                     3                                     3                                       3
                      x 5 + 2x 3                                                                                                                           x3
• D1 =        ∫
              0                   x2 + 1
                                                    dx =            ∫
                                                                    0                   x2 + 1
                                                                                                   dx =    ∫
                                                                                                           0           x2 + 1
                                                                                                                                          dx +     ∫
                                                                                                                                                   0       x2 + 1
                                                                                                                                                                    dx

                  3                                                             3
                                                                                              x dx
              ∫                                                                 ∫x
                              2             2                                           2
          =           x .x x + 1 dx +                                                                    =I+J
                                                                                               2
              0                                                                 0             x +1

                      3                                                                   u = x 2           du = 2x dx
                                                                                                            
                      ∫x                                                                                    ⇒
                                  2             2
Xét I =                               .x x + 1 dx .                                     t                        1 2      32
                      0                                                                   dv = x x 2 + 1 dx  v = ( x + 1)
                                                                                                                3
                                                        3                   3                                                 3
   1 2 2       32                                             2                      32       1          32
I = x ( x + 1)                                                           ∫ x ( x + 1) dx = 8 − ∫ ( x + 1) d ( x + 1)
                                                                                2                   2          2
                                                            −
   3                                                0         3             0
                                                                                                                          3   0

                                                                3
                   2( 2                                                                  2
                                                                                           ( 32 − 1) = 58
                           52
  =8−                x + 1)                                         =8−
                  15                                        0                           15             15

                                                                              u = x 2
                          3
                                           x dx                               
                                                                                             du = 2x dx
                                                                                              
                      ∫x                                                               x dx ⇒ 
                                  2
Xét J =                                                     .               t 
                      0                    x2 + 1                             dv =           v = x 2 + 1
                                                                                              
                                                                              
                                                                                      x2 + 1
                                       3            3                                                3                                                                       3
                                                                                                                                       2 2       32                                  4
                                                ∫ 2x                                                 ∫             + 1 d ( x + 1) = 6 − ( x + 1)
          2           2                                                  2                                    2             2
J=x           x +1 0 −                                              x + 1 dx = 6 −                        x                                                                      =
                                                    0                                                0
                                                                                                                                                       3                 0           3

                                        58 4 26
⇒ D1 = I + J =                            + =
                                        15 3 5
                                                                                                                                      2
                                                                                                                                              1 d ( 1 + x3 )
              2                                                             2                                                                  2
                          1 + x3        1                                                                 = − 1+ x
                                                                                                                    3
• D2 =        ∫              4
                                 dx = −    1 + x3 d  13
                                                                        ∫                                                               +    ∫
              1
                           x            31          x                                                        3 x3                   1       31    x3

                                   2 1 1 d (1 + x )
                                            2                   2                                             2                   3
               2 1 1    x dx
      =         − +
              3 8 2 1 x3 1 + x3
                                =
                                  3
                                    − +     ∫
                                     8 6 1 x3 1 + x3                                                          ∫
                                                            d (u2 )
                                                3                                                              3
               2 1 1                                                                          2 1 1                   du
      =         − +
              3 8 6                             ∫ (u        2
                                                                    − 1) u
                                                                                        =
                                                                                             3
                                                                                               − +
                                                                                                8 3            ∫u     2
                                                                                                                          −1
                                                2                                                              2

                                                                        3
               2 1 1 u −1                                                                    2 1 1 1                  
      =         − + ln                                                              =         − +  ln + 2 ln (1 + 2 ) 
              3 8 3 u +1                                                    2               3  8 3 2                  



220
                                                                                              Bài 8. Phương pháp tích phân t ng ph n

           π 2                                                  π 2
                  sin 2 x                    1
           ∫e                                                      ∫ 2 cos
                                                                                                                                    2
                                         3
• D3 =                      sin x cos x dx =                                         2
                                                                                         x ( 2 sin x cos x ) esin                       x
                                                                                                                                            dx
            0
                                             4                     0

           π2                                                                                                     π2            π2
       1                                                 ) = 1 (1 + cos 2x ) esin                                           1
                 (1 + cos 2x ) d ( esin
                                                 2                                                        2
                                                                                                                                        sin 2 x
            ∫                                                                                                                   ∫e
                                                     x                                                        x
   =                                                                                                                    −                           d (1 + cos 2x )
       4    0
                                                               4                                                  0         4   0

                      π2                                                         π2                                                                 π2
     −1 1                                           1 1                                      1 1 sin                                                          e
                                                                                 ∫ d (e ) = − + e
                             sin 2 x                                                                  2                                     2

                      ∫e
                                                                                       sin x                                                    x
   =   +                               sin 2x dx = − +                                                                                                   =      −1
     2 2              0
                                                    2 2                          0
                                                                                                                        2       2                   0         2

           π 2                                                  π 2
• D4 =     ∫
           π 3
                 cos x ln ( 1 − cos x ) dx =
                                                                π
                                                                 ∫ ln (1 − cos x ) d ( sin x )
                                                                       3

                                                         π2                                                                                 π2
                                          π2                                                                              3              sin x dx
   = sin x ln (1 − cos x )
                                          π3
                                                  −      ∫
                                                         π3
                                                              sin xd ( ln (1 − cos x ) ) =
                                                                                                                         2
                                                                                                                            ln 2 − sin x
                                                                                                                                  π3
                                                                                                                                                ∫
                                                                                                                                         1 − cos x
                            π2            2                                              π2
      3           1 − cos x       3
   =
     2
        ln 2 −
               π3
                  1 − cos x ∫
                            dx =
                                 2
                                    ln 2 − (1 + cos x ) dx
                                          π3
                                                                                         ∫
          3                     π2  3       π       3
   =        ln 2 − ( x + sin x ) =    ln 2 − − 1 +
         2                      π3 2        6      2
         π3                                      π3                                                                                         π3
                                                                                                                                π3
• D5 =
         π4
           ∫ sin x ln( tg x) dx = −π∫ ln ( tg x) d ( cos x) = − cos x ln ( tg x) π
                                                     4
                                                                                                                                    4
                                                                                                                                        +
                                                                                                                                            π4
                                                                                                                                                ∫ cos x d ( ln ( tg x))
                            π3                                                   π3                                              π3
      1            cos x dx     1           dx      1          sin x dx
   = − ln 3 +
      4              2
              π 4 cos x tg x
                             ∫
                             = − ln 3 +
                                4       π4
                                           sin x
                                                 = − ln 3 +
                                                    4              2
                                                            π 4 sin x
                                                                                     ∫                                              ∫
                            π3                                                                                         π3
      1            d ( cos x )    1      1 1 + cos x                                                                                                          3
   = − ln 3 −              2 ∫ = − ln 3 − ln                                                                                = ln (1 + 2 ) −                     ln 3
      4       π 4 1 − cos x
                                  4      2 1 − cos x                                                                   π4                                     4

           π4                            π 4                               π 4                                    π 4                                   π 4
                 x + sin x                                                                                              d (1 + cos x )
                                                                                                                                                         ∫ xd ( tg 2 )
                                                 sin x dx                                x dx                                                                          x
• D6 =     ∫
           0
                 1 + cos x
                           dx =              ∫
                                             0
                                                 1 + cos x
                                                           +               ∫ 2 cos
                                                                            0
                                                                                              2   x
                                                                                                          =−      ∫0
                                                                                                                          1 + cos x
                                                                                                                                       +
                                                                                                                                                         0
                                                                                                  2
                                                         π4        π4                                                                                                  π4
                           x                                                   x           4  π π            x
   =  − ln 1 + cos x + x tg  −
                           20                                        ∫
                                                                       0
                                                                           tg
                                                                                 2
                                                                                   dx = ln     + tg + 2 ln cos
                                                                                           2+ 2 4 8            2                                                       0


                                 π
   = ln
                  4
                             +     ( 2 − 1) + ln 2 + 2 = π ( 2 − 1) + ln1 = π ( 2 − 1)
               2+ 2              4                  4    4                  4


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Chương II: Nguyên hàm và tích phân − Tr n Phương

           π2                                                       π2                                                     π2
• D7 = ∫ sin 2x cos ( sin x ) dx = 2 ∫ sin x cos x cos4 ( sin x) dx = 2 ∫ sin x cos4 ( sin x) d( sin x)
                                         4

            0                                                        0                                                         0

            1                                      1                                        1
                                                 1                       1 (
            ∫                                      ∫                        t 1 + 2 cos 2t + cos 2t ) dt
                                                                                            ∫
                                     4                             2                            2
      = 2 t ( cos t ) dt =                          t (1 + cos 2t ) dt =
            0
                                                 20                      20
                1                                                                      1
           1                   1 + cos 4t       1
      =
           20  ∫
             t  1 + 2 cos 2t +
                                     2
                                            dt =
                                                 40
                                                     t ( 2 + 4 cos 2t + cos 4t ) dt   ∫
                1                        1                                                 2 1             1
        1          1                               t                                                   1                1       
      =
        40      ∫
           2t dt +
                   40                    ∫
                      t ( 4 cos 2t + cos 4t ) dt =
                                                    4                                          0
                                                                                                   +     t d  2 sin 2t + sin 4t 
                                                                                                       40 ∫             4       

                                                                         1        1
       1 1             1          1            1       
      = + t  2 sin 2t + sin 4t  −
       4 4             4
                                       2 sin 2t + sin 4t  dt
                                0 4 0           4                              ∫
                                                                                                                   1
           1 1          1       1             1      
      =     +  2 sin 2 + sin 4  −  − cos 2t − cos 4t 
           4 4          4        4           16      0
           1        1        1       1       31
      =      sin 2 + cos 2 + sin 4 + cos 4 +
           2        4       16      16       64
           π4                                                        π 4                                                       π 4
                     tg x                                                     sin x                                                            2 − cos2 x
• D8 =      ∫
            0
                    cos x
                          1 + sin2 x dx =                                ∫
                                                                         0
                                                                                 2
                                                                             cos x
                                                                                    2 − cos2 x dx = −                              ∫
                                                                                                                                   0
                                                                                                                                                cos2 x
                                                                                                                                                          d ( cos x )


                                                                                                               1       2
            1       2                    2          1       2                                              2                       1       2

      =−        ∫
                1
                        2−u
                             u
                                 2
                                             du =       ∫
                                                        1
                                                                    2−u d 1 =
                                                                          u
                                                                              2−u
                                                                              2

                                                                               u  ()                           1
                                                                                                                               −       ∫
                                                                                                                                       1
                                                                                                                                               1
                                                                                                                                               u
                                                                                                                                                 d   (   2−u
                                                                                                                                                               2
                                                                                                                                                                   )
                             1       2                                                             1   2
                                              du                                           u                                                π
      = 3 −1−                    ∫
                                 1           2−u
                                                    2
                                                        = 3 − 1 − arcsin
                                                                                            2      1
                                                                                                           = 3 −1−
                                                                                                                                           12

           π 3                                                      π 3
                                     x 2 dx                                   x             x cos x
• D9 =      ∫ ( x sin x + cos x )
            0
                                                        2
                                                                =   ∫ cos x ⋅ ( x sin x + cos x )
                                                                     0
                                                                                                                   2
                                                                                                                           dx

      π3                                                                                                   π3          π3
                                                                                                                                                            x 
=−    ∫
      0
             x
                d       1
                         (
           cos x x sin x + cos x
                                 =−
                                      x
                                         ⋅       1
                                    cos x x sin x + cos x 0 )
                                                            +                                                              ∫ x sin x + cos x d  cos x 
                                                                                                                           0
                                                                                                                                               
                                                                                                                                                1
                                                                                                                                                       
                        π3
      −4π                                                       cos x + x sin x  dx = −4π + tg x π 3 = 3 3 − π
                        ∫ x sin x + cos x 
=                   +                        1
                                                                      2                         0
    3+ π 3              0                                          cos x            3+ π 3            3+ π 3



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