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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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