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INTEGRATION
Integration or the process of finding the antiderivative is one of the most important
operations in calculus. It is the opposite process of differentiation.


ANTIDERIVATIVE OR INDEFINITE INTEGRAL


                 is the derivative of the function F x , that is F. x = f x for all x in the
           ` a                                           ` a            ` a   ` a
If     f x
                          ` a                                                           ` a
domain of f, then F x is called the antiderivative or indefinite integral of f x .

                                       We write Z f x dx = F x + C
                                                   ` a         ` a




                                                                                      F x = x 2 @ 2,
                                                                                       ` a
The antiderivative of a given function is not unique. For example,
G x = x 2 + 5, and          H x = x2                                                f x = 2x
     ` a                         ` a                                                ` a
                                            are all antiderivatives of                           as
                            dff       c d b        c d b c
                             ff
                             ff
F. x = G. x = H. x = f x or ff x 2 @ 2 = ff x 2 + 5 = ff x 2 = 2x for all x in the
                                          ff
                                          ff
                                           ff          ff
                                                        ff
                                                       ff
  ` a    ` a    ` a   ` a      b
                            dx           dx           dx
domain of f .


We note that, all the antiderivatives of the function f differ only by a constant (the
derivative of the constant value is always zero). Geometrically, this means that the graphs
           ` a     ` a          ` a
of     F x , G x and H x               are identical except for their vertical position. All their
derivatives will be same for any x = x 0 . All indefinite integrals of f x = 2x are then
                                                                                ` a


expressed by the general form of the antiderivative which is F x = x 2 + C where C is the
                                                                       ` a


constant of integration or an arbitrary constant.
The notation Z is used for integration. The symbol Z f x dx is used to denote the
                                                                    ` a


indefinite integral of the function f(x). The function f(x) is called the integrand.



So, we can write Z 2x dx = x 2 + C .



The indefinite integral of a function is sometimes called the general antiderivative of the
function.

                      INTEGRATION BY INSPECTION
We shall solve the first few questions by method of inspection, that is, comparing the
integrand with some known derivatives.


                                                                f x = sin x
                                                                 ` a
Example : Find the indefinite integral of the function
Solution :
                                                            F x = @ cos x     is F. x = sin x
                                                              ` a                  ` a
        We know that the derivative of the function

        so we write       Z sin x dx = @ cos x + C



                                                                f x = x2
                                                                 ` a
Example : Find the indefinite integral of the function
Solution :
                                                 f 3g
                          dff c
                           ff
                          ff
                          ff 3                 ff xff
                                               dff ff
                                               ff ff
                                                ff f      1f
                                                           ff
                               x = 3x , or               = A 3x 2 = x 2
                            b
                                      2
        We know that
                          dx                   dx    3    3
                                           3
                                          xff
                                           ff
                                            f
                                           ff
        As the derivative of F x =            is F. x = x 2
                                   ` a             ` a
                                           3
                      3
                   xff
                    ff
                     f
        Z x 2 dx = ff+ C
                      3


             STANDARD INTEGRATION FORMULAS:
             COMPARISON WITH DIFFERENTIATIONS
In the above cases of integration by inspection, the antiderivatives were found by
comparing the integrand with some known derivatives. But it is not always possible to do
so, as in most cases the integrand does not match with our known derivatives. So, we
need some integration formulas. As integration is the reverse process of differentiation,
we can make the first few integration formulas directly from the corresponding derivative
formulas. These are given below :


For Polynomial Functions:
Differentiation Formula             Corresponding Integration Formula
                                                  n+1
 dff
 ff`
 ff
  ff                                             xffff
      xn = nA xn@1
                                                  ffff
                                                 ffff
                                      Z x n dx = fffff C
                                                      +
         a
dx                                               n+1
 dff
 ff
  ff
 ff` a
      x = 1                           Z dx = x + C
dx



For Trigonometric Functions:
Differentiation Formula               Corresponding Integration Formula

  dff
  ff
   ff
 ff`
     sin x = cos x                    Z cos x dx = sin x + C
          a
 dx
 dff
 ff
  ff
 ff`
       cos x = @ sin x                Z sin x dx = @ cos x + C
             a
 dx
 dff
 ff
  ff
 ff`
       tan x = sec 2 x                Z sec 2 x dx = tan x + C
             a
 dx
 dff
 ff
  ff
 ff`
       cot x = @csc 2 x               Z csc 2 x dx = @ cot x + C
             a
 dx
 dff
 ff
  ff
 ff`
       sec x = sec x tan x            Z sec x tan x dx = sec x + C
             a
 dx
 dff
 ff
  ff
 ff`
       csc x = @ csc x cot x          Z csc x cot x dx = @ csc x + C
             a
 dx
For Exponential & Logarithmic Functions:
Differentiation Formula                     Corresponding Integration Formula

 dff
 ff
  ff
 ff` x a
     e = ex                                 Z e x dx = e x + C
dx
                                                           x
 dff
 ff` x a
 ff
  ff                                                   afff
     a = a x ln a                                       fff
                                                       fff
                                            Z a x dx = fff+ C a>0, a ≠ 1
dx                                                       ln a
 dff
  ff
 ff`
 ff    a 1f                                   1f
   ln x = fff                                  f
                                               f
                                            Z fdx = ln |x| + C
dx        x                                    x



     INTEGRATION OF COMBINATION OF FUNCTIONS
The following rules of integration for addition, subtraction and scaler multiples of
functions can be used. Note that, unlike differentiation, product and quotients of
functions are not covered in these (which are to be solved using substitution or
integration by parts or by some other methods).



                      Z c f x dx = c Z f x dx
                            ` a                    ` a



                      Z f x F g x dx = Z f x dx F Z g x dx
                          B ` a       ` aC               ` a     ` a




Example : Evaluate Z x 3 dx

Solution :
                    3+1
                  xffff     1f
                   ffff
                    fff      f
                             f
       Z x 3 dx = ffff+ C = fx 4 + C
                    3+1           4


                      1f
                     fff
                     fff
                      fff
                      f
Example : Evaluate Z pwf
                     w dx
                      w
                      w
                      ww
                       w
                       x
Solution :
                                        1
                     1f               @ f+ 1
                                        f
                                        f
          1f         ff
                      ff
                            xfffff       w
                                         ww
                                          w
                                          w
                                          ww
         fff
       Z fwf = Z x @ 2 dx = ffffff C = 2 p x + C
         w dx
         w
          ff
          fff
          f
          w
          ww
           w
                             ff fff
                            ff2fff
                               f
                                  +
         px                   1f
                            @ +1
                              fff
                              2
                            SUBSTITUTION METHOD:
                            CHANGE OF VARIABLES
Till now we have discussed integrals of those functions which are readily obtained from
derivatives of some known functions. They could be solved using the standard formulas.
But some integrals can not be evaluated directly as no standard formula matches the form
                                                               wwww
                                                               wwww
                                                                wwww
                                                                www
                                                                www
                                                                www
                                                                 www
                                                                 www              `     a2
                                                                                     fx ff
                                                                                    lnffff
                                                                                    ffff
                                                                                     ff
                                                                                   fffff
of the given function.          Some examples are          Z x q x 2 + 5 dx ,   Z        dx ,
                                                                                     x

Z tan3 x sec 2 x dx .



One way of solving them is to substitute some new variables in the integral and thus
transforming it to standard form with the new variable. After choosing the suitable
variable, we have to rewrite the integral in terms of the new variable, so that one or more
of the standard formulas can be used. After the integration process is over, the result is to
be written in terms of the original variable.


                                    c4
Example : Evaluate Z 3x x 2 + 1 dx
                            b




Solution :
         Let t = x 2 + 1
                 dtf
                  ff
                   ff
                   f
                  ff
            [         = 2x
                 dx
                         1f
           [              f
                 x dx = fdt
                         2
                     4       3f
                              f
         Z 3x x 2 + 1 dx = Z ft 4 dt
              b         c
                                2
                                   5
                               3f tff
                                f f
                                 f f         3f
                             =                ff
                                             ff
                                 A f C = fft 5 + C
                                     +
                               2 5          10
                                3fb      c5
                                ff
                                 ff
                             = ff x 2 + 1 + C
                               10
Example : Evaluate Z sin 3x dx

Solution :
        Let 3x = t
                     1f
                      ff
             dx =      dt
                     3
                        1f
                         f
        Z sin 3x dx = Z fsin t dt
                            3
                         1f
                          ff           1f
                                        ff
                      = @ cos t + C = @ cos 3x + C
                         3             3


Example : Evaluate Z tan2 2x @ 3 dx
                                `       a


Solution :
       Let 2x @ 3 = t
                 1f
                  ff
           dx = dt
                 2
                            1f
                             f
       Z tan2 2x @ 3 dx = Z ftan2 t dt
             `      a
                                    2
                                             H                   I
                       1f b
                        ff                 1f
                                            ff
                      = Z sec 2 t @ 1 dt = J Z sec 2 t dt @Z dt K
                                     c
                       2                   2
                       1f
                        ff
                      =    tan t @ t + C
                         @          A
                       2
                       1f `
                        ff
                      = tan 2x @ 3 @ 2x @ 3 + C
                         B             a ` aC
                       2


                OTHER INTEGRATION FORMULAS
Apart from the standard formulas given above, the following formulas are also used in
integration (these are obtained by using the substitution method) :
                Z tan x dx = @ ln |cos x| + C


                Z cot x dx = ln |sin x| + C


                Z sec x dx = ln |sec x + tan x| + C


                Z csc x dx = @ ln|csc x + cot x| + C




     dx
  ffffff 1f
   fffff
Z ffffff = farctan f + C
   fffff   ff      xf
                    f
                    f
  x +a
   2    2
           a         a
                L      M
      dx            @ aM
            1ff L xffff
   ffffff
   ffffff
    fffff
Z ffffff = ff lnL fffff + C
            ff
             ff     ffff
                   fffff
                       M
    2    2
           2a L x + a M
                L      M
  x @a
                   + xM
                L     M
    dx
   fffff
   ffffff
  ffffff
  ffffff   ff L affff
           1ff L fffff
            ff
            ff     ffff
                  ffff
Z 2      =    lnL     M+
                      M
                                   C
 a @ x2    2a La @ xM
                           wwww
                           wwww
                           wwww
                            wwww
                            wwww
                            wwww
                            wwww
                           wwww M
     dx
                   L
  ffffffff
   ffffffff
   fffffff
   ffffff
Z fwwwwf = lnLx + q x 2 + a 2 M + C
             L                M
  wwww
   wwww
   wwww
    wwww
    wwww
    wwww
  wwww
  q x2 +
             L                M
           a2
                           wwww M
                           wwww
                           wwww
                            wwww
                            wwww
                            wwww
                            wwww
                           wwwww
     dx
                   L
  ffffffff
  fffffff
   fffffff
    fffff
Z ffwwwff= lnLx + q x 2 @ a 2 M + C
             L                M
  wwww
   wwww
   wwww
   wwww
   wwww
   wwww
    w
  wwwww
 q x2 @ a2
             L                M


     dx           xf
  ffffffff
  fffffff
   fffffff
    fffff
Z ffwwwff= arcsin f+ C
   wwww
   wwww
  wwwww
  wwww
   wwww
   wwww
   wwww
    w
                   f
                   f
 qa 2 @ x 2                a
  wwwww
  wwwww
  wwww
   wwww
   wwww
   wwww
   wwww
  wwwww                    wwwww
                           wwwww
                           wwwww
                           wwww
                            wwww
                            wwww
                            wwww
                            wwww                    wwww
                                                    wwww
                                                    wwww
                                                     wwww
                                                     wwww
                                                     wwww
                                                     wwww
                                                    wwww M
                   1f               1f
                                              L
                    f
                    f                f
                                     f
Z q x 2 + a 2 dx = fx q x 2 + a 2 + fa 2 lnLx + q x 2 + a 2 M + C
                                           L                M
                       2               2   L                M
   wwwww
  wwwww
  wwwww
   wwww
   wwww
   wwww
    wwww
    wwww                   wwwww
                           wwwww
                           wwwww
                           wwww
                            wwww
                            wwww
                            wwww
                            wwww                    wwww M
                                                     wwww
                                                     wwww
                                                     wwww
                                                     wwww
                                                      wwww
                                                      wwww
                                                    wwwww
                       1f q 2     1f 2 L
                                       L
                        f
                        ff         f
                                   ff
Z q x 2 @ a 2 dx   =            2
                         x x @ a @ a lnLx         + q x2 @ a2 M +
                                                              M     C
                       2               2
                                       L                      M
  wwwww
  wwwww
  wwww
   wwww
   wwww
   wwww
   wwww
  wwwww                    wwwww
                           wwwww
                           wwwww
                           wwww
                            wwww
                            wwww
                            wwww
                            wwww
                   1f               1f          xf
                    f
                    f                f
                                     f           f
                                                 f
Z q a 2 @ x 2 dx = fx q a 2 @ x 2 + fa 2 arcsin f + C
                       2               2           a
      dx
  ffffffffff 1f
     ffffff
Z fffwwwwf = farcsec f+ C
  fffffffff
   fffffffff  f
              f      xf
                      f
                      f
    wwww
    wwww
    wwww
     wwww
     wwww
     wwww
    wwwww
  x q x2 @ a2          a       a
                           INTEGRATION BY PARTS
Another useful integration technique for indefinite integrals which do not fit in the basic
formulas is integration by parts. We may consider this method when the integrand is a
product of two functions.


When u and v are differentiable functions of x which is the variable of integration, then
                  d uv = u dv + v du,              u dv = d uv @ v du
                   `   a                                   `   a
                                              or


Integrating both sides we get the following formula for Integration by parts,

                                    Z u dv = uv @ Z v du



While using this method of integration, the given integral is to be separated into two
parts, one part being u and the other part, combined with dx, being dv. For this reason
this method is called integration by parts.
   (a) We have to first choose dv which must be readily integrable to find v. The u
       function will be the remaining part of the integrand that will be differentiated to
       find du.

   (b) The purpose is to find an integral Z v du which is easier to integrate than the


       original integral Z u dv .



Example : Evaluate Z x sec 2 x dx

Solution :
        Let u = x so du = dx
        and dv = sec 2 x dx                 then v =Z dv =Z sec 2 x dx = tan x

        using       Z u dv = uv @ Z v du


        hence           Z x sec 2 x dx = x tan x @Z tan x dx

                                       = x tan x @ @ ln |cos x| + C
                                                      b           c


                                       = x tan x + ln |cos x| + C



Example : Evaluate Z ln x 2 + 4 dx
                             b        c




Solution :
                                                          2x
       Let u = ln x 2 + 4                                fffff
                                                          ffff
                                                         fffff
                                                   du = fffffdx
                    b         c
                                       so
                                                        x +4
                                                          2


       and dv = dx                  then v =Z dv =Z dx = x

       using        Z u dv = uv @ Z v du

                                                          2
                                                 2x ff
                                                fffff
                                                fffff
                                                 ff
             Z ln x 2 + 4 dx = x ln x 2 + 4 @ Z fffffdx
                b        c             b       c
                                                 2
                                                       x +4
                                                             8 G
                                                          fffff
                                                          fffff
                                                           ffff
                                  = x ln x 2 + 4 @ Z 2 @ fffff dx
                                       b        c   F
                                                         x +4
                                                           2


                                                             8 G
                                                          fffff
                                                          fffff
                                                           ffff
                                  = x ln x 2 + 4 @ Z 2 @ fffff dx
                                        b       c   F
                                                         x +4
                                                           2

                                                                 xf
                                                                  f
                                                                  ff
                                  = x ln x 2 + 4 @ 2x + 4 arctan + C
                                        b       c
                                                                 2
                            INTEGRATION USING
                    TRIGONOMETRIC IDENTITIES
When the integrand involves trigonometric functions which can not be solved by the
basic integration formulas, we need to use the different trigonometric identities (together
with substitution if needed) to get them into a form in which the basic integration formula
scan be applied.
The trigonometric identities often used for integration are
           1                   1f             1               ft
                                                            sinff
          ffff
          fff
           fff
sec t = ffff          csc t = fff
                              fff
                               fff
                               f             fff
                                             fff
                                              fff
                                    cot t = ffff             ffff
                                                             fff
                                                    tan t = ffff
         cos t               sin t          tan t           cos t
sin t + cos 2 t = 1
   2


tan2 t + 1 = sec 2 t
cot 2 t + 1 = csc 2 t
            @ cos 2α
          1fffffffff
            ffffffff
            ffffffff
sin α = fffffffff
   2
                2
             + cos 2α
           1ffffffff
             ffffffff
             ffffffff
cos 2 α = fffffffff
                2
                1f
sinαcosα = fsin 2αf
                2
                1f b
                  D                        cE
                 ff
sinαcosβ =          sin α + β + sin α @ β
                               c   b
                2
                1f b
                  D                        cE
                  f
cosαsinβ = f sin α + β @ sin α @ β
                               c   b
                2
                 1f b
                   D                         cE
                  f
                  f
                                 c   b
cosαcosβ =           cos α + β + cos α @ β
                 2
                 1f b
                   D                          cE
                  f
sinαsinβ = f cos α @ β @ cos α + β
                                 c   b
                 2



                        3
Example : Evaluate Z sin x cos 3 x dx

Solution :
        Z sin3 x cos 3 x dx = Z sin3 x cos 2 x A cos xdx


                              = Z sin x 1 @sin x A cos x dx let sin x = t, cos x dx = dt
                                           b            c              B                              C
                                       3            2




                              = Z t 3 @ t 5 dt
                                  b             c


                               1f 4 1f 6
                                f
                                ff    ff
                              =  t @ t +C
                               4     6
                               1f 4
                                f
                                ff       1f 6
                                          ff
                              = sin x @ sin x + C
                               4         6


Example : Evaluate Z se
								
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