Integration Worksheet with Solution - PDF

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					                      Integration Worksheet
                                      Math 132
                                     Solutions

(a) Use partial fractions:
                         x+3           1      10         1   1
                       2 + 5x − 2
                                  dx =             dx +          dx
                    3x                 7    3x − 1       7 x+2
                                       10              1
                                     =    ln(3x − 1) + ln(x + 2) + C.
                                       21              7
(b) Use integration by parts with u = x, dv = e3x dx (which gives du = dx and
        1
    v = 3 e3x ):
                               1                     1               1
                                    x      x                 1
                                    3x
                                       dx = e3x          −               e3x dx
                           0       e       3         0       3   0
                                                     1                   1
                                          x          1
                                        = e3x − e3x
                                          3     0    9                   0
                                                 3
                                          1 + 2e
                                        =          .
                                             9
                               √
(c) Use u substitution. Let u = x + 1 to get u2 = x + 1, 2udu = dx, x = u2 − 1.
    Also, x = 0 ⇒ u = 1 and x = 8 ⇒ u = 3.
                         8 √            u=3
                          x x + 1 dx =      (u2 − 1)(u)(2udu)
                       0                       u=1
                                                                         3
                                              2u3 2u5
                                          =−       +
                                               3     5                   1
                                            1192
                                          =      .
                                             15
                                                                                  1−cos(2t)
(d) Use the “half-angle” formula from trigonometry to write sin2 (t) =               2
                                                                                            .
                                              1 − cos(2t)
                               sin2 (t) dt =              dt
                                                   2
                                           1     1
                                          = t − sin(2t) + C.
                                           2     4
(e) Here, use the fact that cos2 (t) = 1 − sin2 (t):

                                     cos3 (t) dt =       cos(t)(1 − sin2 (t)) dt

                                                               sin3 (t)
                                                  = sin(t) −            + C.
                                                                  3

(f) Here, I recommend using the substitution x = sinh(t). Then, x2 + 1 = sin2 (t) +
    1 = cosh2 (t) and dx = cosh(t)dt. (I prefer the hyperbolic trig substitution here
    to the trigonometric substitution x = tan(t) because, in the end, I think it’s
    harder to deal with sec3 (t) than cosh2 (t).)
        √
            x2 + 1 dx =              cosh2 (t) cosh(t) dt

                     =             cosh2 (t) dt
                            e2x + 2 + e−2x
                     =                     dt
                                   4
                       e2t + 4t − e−2t
                     =                 +C
                              8

    To finish, and put things in terms of x again, use the fact that x = sinh(t) ⇒
               √
    t = ln(x + 1 + x2 ) and a little algebra:


                                     √                          √                              √
                              (x +       1 + x2 )2 + 4 ln(x +       1 + x2 ) − (x +                1 + x2 )−2
                     =                                                                                          +C
                        √             √     8
                       x 1 + x2 ln(x + 1 + x2 )
                     =         +                + C.
                          2           2

(g) Use integration by parts with u = arcsin(x) and dv = dx (then we have du =
    √ dx  and v = x).
     1−x2

                              1                                     1           1
                                                                                        x dx
                                  arcsin(x) dx = x arcsin(x)            −           √
                          0                                         0       0           1 − x2
                                                                    1       √              1
                                                  = x arcsin(x)         +       1 − x2
                                                                    0                      0
                                                   π
                                                  = − 1.
                                                   2


                                                     2
(h) Here, we use parts with u = (ln(x))2 and dv = dx (then we have du = 2 ln(x) dx
                                                                                x
    and v = x).
                           e                                      e               e
                                        2                     2
                               (ln(x)) dx = x(ln(x))                  −               2 ln(x) dx
                       1                                          1           1
                                                                  e                                e
                                                              2
                                                = x(ln(x))            + 2x − 2x ln(x)
                                                                  1                                1
                                                = e − 2.

                                            1
(i) Here use the fact that ln x 2                = 1 ln(x):
                                                   2

                                            √           1
                                   ln           x dx =    ln(x) dx
                                                        2
                                                       x x ln(x)
                                                     =− +          + C.
                                                       2       2

(j) Here you have some choices. You could use the substitution x = 2 cosh(t).
    This time, we’ll use the trigonometric substitution x = 2 sec(t). This leads
       √
    to x2 − 4 = 2 tan(t) and dx = 2 sec(t) tan(t)dt. Let’s ignore the limits of
                    √
                     20   dx
    integration in √5 x2 √x2 −4 at first.

                                      dx                      2 sec(t) tan(t) dt
                                     √       =
                                   x2 x2 − 4                   8 sec2 (t) tan(t)
                                                      1
                                                     =    cos(t) dt
                                                      4
                                                      1
                                                     = sin(t) + C.
                                                      4
    Probably the tidiest way to finish, is to use the fact that when x = 2 sec(t) =
    √                                      √
                    1                                         2
      5 ⇒ sin(t) = √5 and x = 2 sec(t) = 20 ⇒ sin(t) = √5 . Then, the definite
    integral
                √                                          √
              x= 20                                      x= 20
                         dx      1                                        1           2   1              1
                        √       = sin(t)                              =               √ −√             = √ .
               √
             x= 5     x2 x2 − 4  4                         √
                                                         x= 5             4            5   5            4 5




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