Integration by Substitution Examples with Explanation and PDF

The Method of substitution is a very important part of Calculus. Substitution is the counterpart to the chain rule for differentiation. Here we are going to solve the problem using the substitution method with a step-by-step explanation. We are also providing PDFs for offline practice.

Problem 1:

`\int \frac { ( 1 + x ) e ^ { x } } { \cos ^ { 2 } ( x e ^ { x } ) } d x`

Solution:

Let,

`x e ^ { x }=u`
`\Rightarrow \frac { d } { d x } ( x e ^ { x } ) = \frac { d u } { d x }`
`\Rightarrow e ^ { x } \frac { d } { d x } ( x ) + x \frac { d } { d x } ( e ^ { x } ) = \frac { d u } { d x }`

[Note: here we used the formula `\frac { d } { d x } ( u v ) = u \frac { d v } { d x } + v \frac { d u } { d x }`]

`\Rightarrow e ^ { x } + x e ^ { x } = \frac { d u } { d x }`

[Note: here, `\frac { d } { d x } ( x ) = 1` and `\frac { d} { d x } ( e ^ { x } ) = e ^ { x }`]

`\Rightarrow ( 1 + x ) e ^ { x } = \frac { d u } { d x }`
`\Rightarrow ( 1 + x ) e ^ { x } \cdot d x = d u`

Now,

`\int \frac { ( 1 + x ) e ^ { x } } { \cos ^ { 2 } ( x e ^ { x } ) } d x`

`= \int \frac { ( 1 + x ) e ^ { x } \cdot d x } { \cos ^ { 2 } ( x e ^ { x } ) }`
`= \int \frac { d u } { \cos ^ { 2 } u }`
`= \int \sec ^ { 2 } u d u`
`= \tan u + c`

[Note: Formula: `\int \sec ^ { 2 } x d x = \tan x + c`]

`\int \sec ^ { 2 } x d x = \tan x + c`

Answer: `\int \sec ^ { 2 } x d x = \tan x + c`

Here is the handwritten solution of this math:

Problem 2:

`\int \frac { e ^ { m \tan ^ { - 1 } x } } { 1 + x ^ { 2 } } d x`

Solution:

let,

`m \tan ^ { - 1 } x = u`
`\Rightarrow m \frac { 1 } { 1 + x ^ { 2 } } = \frac { d u } { d x }`

[Note: Formula used here, `\frac { d } { d x } ( \tan ^ { - 1 } x ) = \frac { 1 } { 1 + x ^ { 2 } }`]

`\Rightarrow \frac { m } { 1 + x ^ { 2 } } d x = d u`
`\Rightarrow  \frac { 1 } { 1 + x ^ { 2 } } d x = \frac { 1 } { m } d u`

Now,

`\int \frac { e ^ { m \tan ^ { - 1 } x } } { 1 + x ^ { 2 } } d x`
`= \int \frac { 1 \cdot d x } { 1 + x ^ { 2 } } ( e ^ { mtan - 1 } x )`
`= \frac { 1 } { m } \int e ^ { u } d u`
`= \frac { 1 } { m } e ^ {u} + c`
`= \frac { 1 } { m } e ^ { mtan ^ { - 1 } x } + c`

Answer: `\frac { 1 } { m } e ^ { mtan ^ { - 1 } x } + c`

Here is the handwritten solution of this math:

You can find other Math Problems with the proper solutions in the PDF file. Download PDF and try them to understand. Comment below if you face any problem understanding.