Question

Solve the following integral.

$\int {\sin }^{6}xdx$

Short answer

Answer is$\frac{-15\sin 2x}{64}+\frac{3\sin 4x}{64}-\frac{\sin 6x}{96}+\frac{5}{16}x+C$

Step-by-step solution

Step 1. Given information

An integral is$\int {\sin }^{6}xdx$

Step 2. Explanation

$\begin{array}{rcl}\int {\sin }^{6}xdx& =& \int \frac{-15\cos 2x}{32}+\frac{3\cos 4x}{16}-\frac{\cos 6x}{32}+\frac{5}{16}dx\\ & =& \frac{-15\sin 2x}{64}+\frac{3\sin 4x}{64}-\frac{\sin 6x}{96}+\frac{5}{16}x+C\end{array}$