# Integrals Involving $$\sin(x)$$ with Even Power

Tutorial to find integrals involving even powers of $$\sin(x)$$, using reducing power formulas, are presented. Exercises with answers are at the bottom of the page. Integrating odd powers of sine is much simpler.

## Review Power Reducing Formulas

For $$n$$ a positive integer, we have the following formulas in trigonometry that may be used to reduce the power of $$\sin(x)$$.
(a) $$\sin^2(x) = \dfrac{1}{2}(1 - \cos(2x))$$
(b) $$\sin^3(x) = \dfrac{1}{4}(3\sin(x) - \sin(3x))$$
(c) $$\sin^4(x) = \dfrac{1}{8}(3 - 4\cos(2x) + \cos(4x))$$
(d) $$\sin^5(x) = \dfrac{1}{16}(\sin(5x) - 5\sin(3x) + 10\sin(x))$$
(e) $$\sin^6(x) = \dfrac{1}{32}(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x))$$

## Examples with Detailed Solutions

In what follows, $$C$$ is the constant of integration.

### Example 1

Evaluate the integral $\int \sin^2(x) \, dx$ Solution to Example 1:
The main idea is to use the identity $$\sin^2(x) = \dfrac{1}{2}(1 - \cos(2x))$$ to reduce the power and hence make the integral easily evaluated as follows:
$$\displaystyle \int \sin^2(x) \, dx = \int \dfrac{1}{2}(1 - \cos(2x)) \, dx$$
$$= \displaystyle \dfrac{1}{2}\int dx - \dfrac{1}{2}\int \cos(2x) \, dx$$
$$= \dfrac{1}{2}x - \dfrac{1}{4}\sin(2x) + C$$

### Example 2

Evaluate the integral $\int [\sin^2(x) - 16\sin^6(x)] \, dx$ Solution to Example 2:
Use the power reducing formulas to rewrite the integral as follows
$$\displaystyle \int [\sin^2(x) - 16\sin^6(x)] \, dx$$
$$\displaystyle = \int \left[\dfrac{1}{2}(1 - \cos(2x)) - 16\left(\dfrac{1}{32}\right)(10 - 15\cos(2x) + 6\cos(4x) - \cos(6x))\right] \, dx$$
$$\displaystyle = \dfrac{1}{2} \int [1 - \cos(2x) - 10 + 15\cos(2x) - 6\cos(4x) + \cos(6x)] \, dx$$
$$\displaystyle = \dfrac{1}{2} \int [-9 + 14\cos(2x) - 6\cos(4x) + \cos(6x)] \, dx$$
$$\displaystyle = \dfrac{1}{2}[-9x + 7\sin(2x) - \dfrac{3}{2}\sin(4x) + \dfrac{1}{6}\sin(6x)] + C$$

## Exercises

Evaluate the following integrals.
(a) $$\displaystyle \int [8\sin^6(x) - 2\sin^2(x)] \, dx$$
(b) $$\displaystyle \int [4\sin^4(x) + \sin^2(x)] \, dx$$

(a) $$\dfrac{3}{2}x - \dfrac{11}{8}\sin(2x) + \dfrac{3}{8}\sin(4x) - \dfrac{1}{24}\sin(6x) + C$$
(b) $$2x - \dfrac{5}{4}\sin(2x) + \dfrac{1}{8}\sin(4x) + C$$