# Integral by Substitution: Indefinite Integral

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

## Example∫ (2x - 1)^{8} dx

Instead of expanding (2x - 1)^{8} by hand,

let's solve this integral by substitution.

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) = ∫ t^{8} [1/2]dt.

Take 1/2 out from the integral.

Solve the integral.

Write the coefficient 1/2.

The integral of x^{8} is [1/9]x^{9}.

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∫ sin^{5} x cos x dx

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:

∫ sin^{5} x cos x dx.

Then (given) = ∫ t^{5} dt.

The integral of x^{5} is [1/6]x^{6}.

This is an indefinite integral.

So write +C.

sin x = t

So change t back to sin x.

So

[1/6]sin^{6} x + C

is the answer.