We have already discussed the problems on integration of functions in standard forms and the problems involving combinations of these functions. Integrals of certain functions can not be obtained directly if they are not in one of the standard forms, but may be reduced to standard forms by proper substitution. The method of evaluating an integral by reducing it into standard form by a proper substitution is called integration by substitution.The method of substitution is very much essential while integrating composite functions.
Let us consider two functions such that g(x) = f(h(x)), where h(x) is a continuously differentiable function.
Then to evaluate $\int$ f(h(x)) . h'(x). dx,
we substitute U = h ( x )
then h'(x) dx = dU
Substituting this in the above integral we get,

$\int$ f(h(x)) . h'(x). dx = $\int$ f(U) . dU

After evaluating this integral we substitute back the value of U.

## U Substitution Formula

The following formulae are derived by using U substitution.

1. Evaluate $\int$ ( a x + b )n dx

Proof: Let U = ax + b,

=> dU = a . dx

=> dx = $\frac{dU}{a}$

$\int$ ( ax + b )n dx = $\int$ Un . $\frac{dU}{a}$

= $\frac{U^{n+1}}{a(n+1)}$ + C

= $\int$ $\frac{(ax+b)^{n+1}}{a(n+1)}$ + C, where n $\ne$ -1

2. Evaluate $\int$ eax+b dx

Proof : Let U = ax + b

=> dU = a . dx

=> dx = $\frac{dU}{a}$

$\int$ eax+b dx = $\int$ eU . $\frac{dU}{a}$

= $\frac{1}{a}$ $\int$ eU . dU

= $\frac{1}{a}$ eU

= $\frac{1}{a}$ e(ax+b) + C

Following above procedures we have the following formulae by U substitution.

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

4. $\int$ abx+c dx = $\frac{1}{b}$ $\frac{a^{bx+c}}{log\;a}$ + C

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

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

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

8. $\int$ cossec2 (ax + b ) dx = - $\frac{1}{a}$ cot ( ax + b ) + C

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

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

## Integration by U Substitution

### We follow the following steps while integrating by U Substitution :

1. We first identify the function and its derivative in the given integral.
2. Assume the function as U and its derivative along with dx as dU.
3. When the integral is a function in terms of U, integrate by standard methods.
4. Substitute U as the function in Step 2 and express the answer in terms of x.

### 1. Integrals of the form $\int$ $\frac{f'(x)}{f(x)}$. dx

Substituting U = f ( x ), we get, f ' ( x ) dx = dU
Therefore the given integral will become
$\int$ $\frac{dU}{U}$ = ln{U} + C

= ln {f(x)| + C

### 2. Integrals of the form $\int$ f(x). f'(x). dx

Substituting U = f( x), we get f ' (x) dx = dU
Therefore, the given integral will become,
$\int$ U dU = $\frac{U^{2}}{2}$ + C

= $\frac{(f(x))^{2}}{2}$ + C [ substituting U = f ( x ) ]

### Solved Examples

Question 1: Evaluate $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx
Solution:

Let I = $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx

Substitute U = x2 + x + 3
Differentiating both sides, we get
dU = ( 2x + 1 ) dx
Putting in the given integral I we get,
I = $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx

= $\int$ $\sqrt{U}$ dU

= $\frac{U^{\frac{1}{2}+1}}{\frac{1}{2}+1}$

=  $\frac{U^{\frac{3}{2}}}{\frac{3}{2}}$

= 2  $\frac{U^{\frac{3}{2}}}{3}$

= $\frac{2}{3}$ $( x^{2}+x+3)^{\frac{3}{2}}$ [ substituting U = f ( x ) ]

Therefore, $\int$ $\sqrt{x^{2}+x+3}$ (2x + 1 ) dx = $\frac{2}{3}$ $(x^{2}+x+3)^{\frac{3}{2}}$ + C

Question 2: Evaluate $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx
Solution:

We have, I = $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx

Substituting,  U = ex + e-x ,

we get, dU = ( ex - e-x ) dx

Substituting in the given integral,

we get,                   I = $\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ dx

= $\int$ $\frac{dU}{U}$

= ln | U | + C

= ln | ex + e-x | + C

$\int$ $\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$ = ln | ex + e-x | + C

## U Substitution Examples

### Solved Examples

Question 1: Evaluate $\int$   $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx
Solution:

We have I = $\int$   $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx

Let U = cos x + sinx

Therefore,            dU = ( - sin x + cos x ) dx

Substituting in the given integral we get,

I = $\frac{dU}{U}$

= ln | U | + C

= ln | cos x + sin x | + C

$\int$   $\frac{cos\;x-sin\;x}{cos\;x+sin\;x}$ dx = ln | cos x + sin x | + C

Question 2: Evaluate $\int$ cot x . dx
Solution:

We have I = $\int$ cot x . dx

= $\int$ $\frac{cos\;x}{sin\;x}$ dx

Let U = sin x, we get dU = cos x dx

Therefore,    I = $\int$ $\frac{cos\;x}{sin\;x}$ dx

= $\int$ $\frac{dU}{U}$

= ln | U | + C

= ln | sin x | + C

$\int$ cot x . dx  = ln | sin x | + C

Question 3: Evaluate $\int$ sec x . dx
Solution:

We have I = $\int$ sec x . dx

Multiplying the numerator and the denominator by ( sec x + tan x ), we get,

I = $\int$ sec x . $\frac{sec\;x+tan\;x}{sec\;x+tan\;x}$ dx

Let U = sec x + tan x
Therefore,          dU = ( sec x. tan x + sec2 x ) dx
= sec x ( tanx + sec x ) dx
Substituting in the given integral, we get,

I = $\int$ sec x . $\frac{sec\;x+tan\;x}{sec\;x+tan\;x}$ dx

= $\int$ $\frac{dU}{U}$

= ln | U | + C

= ln | sec x + tan x | + C

$\int$ sec x . dx  = ln | sec x + tan x | + C

Question 4: Evaluate $\int$ sin3 x dx
Solution:

Let     I = $\int$ sin3 x dx

= $\int$ sin2 x . sinx . dx

= $\int$ ( 1 - cos2 x ) sin x . dx
Let  U = cos x
we get, dU = - sin x . dx
=> sin x . dx = - dU
Substituting in the given integral, we get
I = $\int$ ( 1 - cos2 x ) sin x . dx

= $\int$ ( 1 - U2 ) ( - dU )

= $\int$ ( U2 -1 )  dU

= $\frac{U^{3}}{3}$ - U + C

= $\frac{sin^{3}x}{3}$ - sin x + C

$\int$ sin3 x dx  = $\frac{sin^{3}x}{3}$ - sin x + C.