# 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. $$\dfrac{1}{17}\cos^{17}(x) - \dfrac{1}{15}\cos^{15}(x)$$

### More references and Links

integrals and their applications in calculus.