# Integrals Involving sin(x) and cos(x) with Odd Power

Tutorial to find integrals involving the product of powers of \( \sin(x) \) and \( \cos(x) \) with one of the two having an odd power. Examples and exercises with solutions are included.

## Examples with Detailed solutions

In what follows, C is the constant of integration.

### Example 1

Evaluate the integral \[ \int \sin^3(x) \cos^2(x) dx \]__Solution to Example 1:__

The main idea is to rewrite the integral writing the term with the odd power as the product of a term with power 1 and a term with an even power.

Example: \( \sin^3(x) = \sin^2(x) \sin(x) \).

Hence the given integral may be written as follows:

\( \displaystyle \int \sin^3(x) \cos^2(x) dx = \int \sin^2(x) \cos^2(x) \sin(x) dx \)

We now use the identity \( \sin^2(x) = 1 - \cos^2(x) \) and rewrite the given integral as follows:

\( \displaystyle \int \sin^3(x) \cos^2(x) dx = \int (1 - \cos^2(x)) \cos^2(x) \sin(x) dx \)

We now let \( u = \cos(x) \), hence \( du/dx = -\sin(x) \) or \( -du = \sin(x)dx \) and substitute in the given integral to obtain

\( \displaystyle \int \sin^3(x) \cos^2(x) dx = -\int (1 - u^2) u^2 du \)

Expand and calculate the integral on the right

\( \displaystyle \int \sin^3(x) \cos^2(x) dx = \int (u^4 - u^2) du \)

\( = \dfrac{1}{5}u^5 - \dfrac{1}{3}u^3 + C \)

Substitute \( u \) by \( \cos(x) \) to obtain

\[ \displaystyle \int \sin^3(x) \cos^2(x) dx = \dfrac{1}{5}\cos^5(x) - \dfrac{1}{3}\cos^3(x) + C \]

### Example 2

Evaluate the integral\[ \int \sin^{12}(x) \cos^5(x) dx \]

__Solution to Example 2:__

Rewrite \( \cos^5(x) \) as follows \( \cos^5(x) = \cos^4(x) \cos(x) \).

Hence the given integral may be written as follows:

\( \displaystyle \int \sin^{12}(x) \cos^5(x) dx = \int \sin^{12}(x) \cos^4(x) \cos(x) dx \)

We now use the identity \( \cos^2(x) = 1 - \sin^2(x) \) to rewrite \( \cos^4(x) \) in terms of power of \( \sin(x) \) and rewrite the given integral as follows:

\( \displaystyle \int \sin^{12}(x) \cos^5(x) dx = \int \sin^{12}(x) (1 - \sin^2(x))^2 \cos(x) dx \)

We now let \( u = \sin(x) \), hence \( du/dx = \cos(x) \) or \( du = \cos(x)dx \) and substitute in the given integral to obtain

\( \displaystyle \int \sin^{12}(x) \cos^5(x) dx = \int u^{12} (1 - u^2)^2 du \)

Expand and calculate the integral on the right

\( \displaystyle \int \sin^{12}(x) \cos^5(x) dx = \int (u^{16} - 2u^{14} + u^{12}) du \)

\( = \dfrac{1}{17}u^{17} - \dfrac{2}{15}u^{15} + \dfrac{1}{13}u^{13} + C \)

Substitute \( u \) by \( \sin(x) \) to obtain

\( \displaystyle \int \sin^{12}(x) \cos^5(x) dx = \dfrac{1}{17}\sin^{17}(x) - \dfrac{2}{15}\sin^{15}(x) + \dfrac{1}{13}\sin^{13}(x) + C \)

## Exercises

Evaluate the following integrals.1. \( \displaystyle \int \cos^3(x) \sin^2(x) dx \)

2. \( \displaystyle \int \sin^3(x) \cos^{14}(x) dx \)

### Answers to Above Exercises

1. \( -\dfrac{1}{5}\sin^5(x) + \dfrac{1}{3}\sin^3(x) \)2. \( \dfrac{1}{17}\cos^{17}(x) - \dfrac{1}{15}\cos^{15}(x) \)