Question

Solve the following integral.

$\int {\sin }^{5}x{\cos }^{2}xdx$

Short answer

Answer is$\frac{-{\cos }^{7}x}{7}+\frac{2{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+C$

Step-by-step solution

Step 1. Given information

An integral is$\int {\sin }^{5}x{\cos }^{2}xdx$

Step 2. Explanation

$\int {\sin }^{5}x{\cos }^{2}xdx=\int {\left(1-{\cos }^{2}x\right)}^2\sin x{\cos }^{2}xdx$

Let $u=\cos x$

$\begin{array}{rcl}\int {\sin }^{5}x{\cos }^{2}xdx& =& -\int u^6-2u^4+u^{2}du\\ & =& \frac{-u^7}{7}+\frac{2u^5}{5}-\frac{u^3}{3}+C\\ & =& \frac{-{\cos }^{7}x}{7}+\frac{2{\cos }^{5}x}{5}-\frac{{\cos }^{3}x}{3}+C\end{array}$