The term "Integration" is used to find the function whose derivative is given, which is called the anti derivative of a function. For instance, if we know the acceleration of an object at any instant, then there exists a question, whether we can find the velocity or the position of the object at any instant. There are several such practical situations where the process of integration is involved.
The development of integral calculus arises out of the efforts of solving the problems of the following types.

1. The problem of finding the function when its derivative is given.
2. The problem of finding the area bounded by the graph of a function under certain conditions.
These two types lead to two forms of integrals, indefinite and definite integrals which together constitute Integral Calculus. In this section we shall discuss about the method of integration by substitution.

## Integration by Substitution Formula

By the method of substitution we change the given integral of one variable into another integral of another variable.
Let the integral be I = $\int$ f(x) dx

Substituting x = g(t),

$\frac{dx}{dt}$ = g'(t)

=> dx = g'(t) . dt

Substituting this in the given integral we get,

I = $\int$ f(x) . dx

= $\int$ f (g (t)) . g' (t) . dt

The change of variable formula is one of the important tools available to us in the name of integration by substitution.
Let us discuss with some of the Formulae to integrate by substitution.

1. $\int$ f (ax + b) dx = $\frac{1}{a}$ f (ax + b)

2. $\int$ sin (ax + b) dx = - $\frac{1}{a}$ cos (ax + b) + C

3. $\int$ cos (ax + b) dx = $\frac{1}{a}$ sin (ax + b) + C

4. $\int$ csc2(ax + b) dx = - $\frac{1}{a}$ cot (ax + b) + C

5. $\int$ sec2(ax + b) dx = - $\frac{1}{a}$ tan (ax + b) + C

6. $\int$ sec (ax + b) . tan (ax + b) dx = $\frac{1}{a}$ sec (ax + b) + C

7. $\int$ csc (ax + b) . cot (ax + b) dx = - $\frac{1}{a}$ csc (ax + b) + C

8. $\int$ tan (ax + b) dx = - $\frac{1}{a}$ ln | cos (ax + b) | + C

9. $\int$ cot (ax + b) dx = - $\frac{1}{a}$ ln | sin (ax + b) | + C

10. $\int$ sec (ax + b) dx = - $\frac{1}{a}$ ln | sec (ax + b) + tan (ax + b) | + C

11. $\int$ csc (ax + b) dx = - $\frac{1}{a}$ ln | csc (ax + b) - cot (ax + b) | + C

12. $\int$ $\frac{1}{ax+b}$ dx = $\frac{1}{a}$ ln | ax + b | + C

13. $\int$ eax+b dx = $\frac{1}{a}$ eax+b + C

14. $\int$ abx+c dx = $\frac{1}{b}$ $\frac{a^{bx+c}}{ln|a|}$ + C

## Integration by Substitution Examples

### Solved Examples

Question 1: Evaluate $\int$ x $\sqrt{(x^{2}\;-\;5)}$
Solution:

We have,            I = $\int$ x $\sqrt{(x^{2}\;-\;5)}$
Substituting, u = x2 - 5

we get        $\frac{du}{dx}$ = 2x

=> du = 2 x . dx

=>  $\frac{du}{2}$ = x dx

Substituting in the given integral I, we get,
I = $\int$   $\sqrt{x^{2}-5}$ x dx

= $\int$ $\sqrt{u}$ $\frac{du}{2}$

= $\frac{1}{2}$ $\int$ $\sqrt{u}$  du

= $\frac{1}{2}$ . $\frac{u^{(\frac{1}{2}+1)}}{\frac{1}{2}+1}$ + C

= $\frac{1}{2}$ . $\frac{u^{3/2}}{\frac{1}{2}}$ + C

$\int$ x $\sqrt{x^{2}\;-\;5}$
= $\frac{1}{2}$ . $\frac{(x^{2}\;-\;5)^{\frac{3}{2}}}{\frac{1}{2}}$ + C

Question 2: Evaluate $\int$ x ( 1 + x ) 5 . dx
Solution:

We have I = $\int$ x ( 1 + x ) 5 . dx
Substituting, u = 1 + x

we get $\frac{du}{dx}$ = 1

=> du = dx
and 1 + x = u
=>   x = u - 1
Therefore, the given integral will become,
I = $\int$ ( u - 1 ) u5 . du
= $\int$ ( u6 - u5 ) . du

= $\frac{u^{7}}{7}$ - $\frac{u^{6}}{6}$ + C

$\int$ x ( 1 + x ) 5 . dx  = $\frac{(1\;+\;x)^{7}}{7}$ - $\frac{(1\;+\;x)^{6}}{6}$ + C

Question 3: Evaluate $\int$ $\frac{ln\;|\;x\;|\;}{x}$ dx
Solution:

We have I = $\int$ $\frac{ln|x|}{x}$ dx

Substituting, u = ln|x|

we get, $\frac{du}{dx}$ = $\frac{1}{x}$

Therefore, du = $\frac{dx}{x}$

Substituting in the given integral we get,

I = $\int$ $\frac{ln|x|}{x}$ dx

= $\int$ u du

= $\frac{u^{2}}{2}$ + C

$\int$ $\frac{ln\;|\;x\;|\;}{x}$ dx  = $\frac{(ln\;|\;x\;|\;)^{2}}{2}$ + C

Question 4: $\int$ $\frac{sinx}{2\;+\;3cosx}$ dx
Solution:

We have I = $\int$ $\frac{sinx}{2+3cosx}$ dx

Substituting, u = 2 + 3 cos x

we get, $\frac{du}{dx}$ = - 3 sin x

Therefore, we have d u = - 3 sin x. dx

=>   $\frac{du}{-3}$ = sin x. dx

Substituting in the given integral, we get,

I = $\int$ $\frac{sinx}{2+3cosx}$ dx

= $\int$ $\frac{1}{u}$ . $\frac{du}{-3}$

= $\frac{1}{-3}$ $\int$ $\frac{du}{u}$

= - $\frac{1}{3}$ . ln | u | + C

$\int$ $\frac{sinx}{2\;+\;3cosx}$ dx       = - $\frac{1}{3}$ . ln | (2 + 3 cos x ) | + C

Question 5: $\int$ $\frac{3x^{2}}{1\;+\;x^{6}}$ dx
Solution:

We have, I = $\int$ $\frac{3x^{2}}{1+x^{6}}$ dx

Substituting, u = x3

We get, $\frac{du}{dx}$ = 3 x2 .

=>                    du = 3 x2 . dx
Since u = x3
we have u2 = ( x3 )2 = x6

Substituting, in the given integral we get,
I = $\int$ $\frac{3x^{2}}{1\;+\;x^{6}}$ dx

= $\int$ $\frac{du}{1\;+\;u^{2}}$

= arc tan u + C

= arc tan ( 1 + x3 ) + C
therefore, we get,
$\int$ $\frac{3x^{2}}{1\;+\;x^{6}}$ dx = arc tan ( 1 + x3 ) + C

Question 6: $\int$ sec4 x . dx
Solution:

We have I = $\int$ sec4 x . dx
= $\int$ sec2 x . sec2 x . dx
= $\int$ ( 1 + tan2 x ) . sec2 x . dx
Substituting u = tan x

$\frac{du}{dx}$ = sec2 x

=> du = sec2 x . dx
Substituting in the given integral, we get,
I = $\int$ sec4 x . dx
= $\int$ ( 1 + tan2 x ) . sec2 x . dx
= $\int$ ( 1 + u2 ) du
= u + $\frac{u^{3}}{3}$  + C

$\int$ sec4 x . dx   = tan x + $\frac{tan^{3}x}{3}$  + C

## Integration by Substitution Practice

### Practice Problems

Question 1: $\int$ (4 - 9 x)5 . dx
Question 2: $\int$ sin5 x dx
Question 3: $\int$ $\frac{x^{3}}{x^{2}+1}$ dx
Question 4: $\int$ $\frac{(x-1)}{\sqrt{x-4}}$ dx
Question 5: $\int$ $\frac{2x+5}{x^{2}+5x+9}$ dx
Question 6: $\int$ $\frac{secx}{ln|secx\;+\;tanx|}$ . dx
Question 7: $\int$ $\frac{1\;+\;cosx}{1\;-\;cosx}$ . dx
Question 8: $\int$ $\frac{(x+1)}{x^{2}+2x-3}$
Question 9: $\int$ etan x . sec2 x. dx
Question 10: $\int$ $\frac{1}{e^{x}\;+\;e^{-x}}$ . dx