# Integral by Substitution: Indefinite Integral

How to solve the given indefinite integral by using the integral by substitution: 2 examples and their solutions.

## Example 1

### Example

Instead of expanding (2x - 1)8 by hand,
let's solve this integral by substitution.

### Solution

Set 2x - 1 = t.

Differentiate both sides.

The derivative of 2x - 1 is 2dx.
And the derivative of t is dt.

Derivative of an Implicit Function

To make dx in the given integral,
change 2dx = dt
to dx = [1/2]dt.

2x - 1 = t
dx = [1/2]dt

Put these into ∫ (2x - 1)8 dx.

Then (given) = ∫ t8 [1/2]dt.

Take 1/2 out from the integral.

Solve the integral.

Write the coefficient 1/2.

The integral of x8 is [1/9]x9.

Integral of a Polynomial

This is an indefinite integral.
So write +C.

[1/2]⋅[1/9] = 1/18

2x - 1 = t
So change t back to (2x - 1).

So [1/18](2x - 1)9 + C is the answer.

## Example 2

### Solution

Set sin x = t.

Differentiate both sides.

The derivative of sin x is cos x dx.

Derivative of sin x

And the derivative of t is dt.

cos x dx is already in the given integral.
So you don't have to change cos x dx = dt.

sin x = t
cos x dx = dt

Put these into the given integral:
∫ sin5 x cos x dx.

Then (given) = ∫ t5 dt.

The integral of x5 is [1/6]x6.

This is an indefinite integral.
So write +C.

sin x = t
So change t back to sin x.

So [1/6]sin6 x + C is the answer.