Question

Consider the integral $\int {\sin }^{2}x{\cos }^{2}xdx$. We solved this integral in Example 2(b) by applying double-angle identities at the very beginning.

(a) Solve this integral by applying the identity ${\sin }^{2}x=1-{\cos }^{2}x$

(b) Solve this integral another way, by applying the identity$\sin x\cos x=\frac{1}{2}\sin 2x$

Short answer

(a)$\begin{array}{rcl}& & \frac{x}{8}-\frac{\sin 4x}{32}+C\end{array}$

(b)$\begin{array}{rcl}& & \frac{x}{8}-\frac{\sin 4x}{32}+C\end{array}$

Step-by-step solution

Part (a) Step 1. Explanation

$\begin{array}{rcl}\int {\sin }^{2}x{\cos }^{2}xdx& =& \int \left(1-{\cos }^{2}x\right){\cos }^{2}xdx\\ & =& \int {\cos }^{2}x-{\cos }^{4}xdx\\ & =& \int \frac{1}{2}+\frac{\cos 2x}{2}-{\cos }^{4}xdx\\ & =& \frac{x}{2}+\frac{\sin 2x}{4}-\int {\cos }^{4}xdx\\ & =& \frac{x}{2}+\frac{\sin 2x}{4}-\int \frac{\cos 2x}{2}+\frac{\cos 4x}{8}+\frac{3}{8}dx\\ & =& \frac{x}{2}+\frac{\sin 2x}{4}-\frac{\sin 2x}{4}-\frac{\sin 4x}{32}-\frac{3}{8}x+C\\ & =& \frac{x}{8}-\frac{\sin 4x}{32}+C\end{array}$

Part (b) Step 1. Explanation

$\begin{array}{rcl}\int {\sin }^{2}x{\cos }^{2}xdx& =& \int \frac{1}{4}{\sin }^{2}2x\\ & =& \frac{1}{4}\int \frac{1}{2}-\frac{\cos \left(4x\right)}{2}dx\\ & =& \frac{x}{8}-\frac{\sin 4x}{32}+C\end{array}$