Example 5. Find the integral $\int \sin ^{4} x \cos ^{2} x d x$.
## Solution.

$$
\begin{gathered}
\int \sin ^{4} x \cos ^{2} x d x=\int(\sin x \cos x)^{2} \sin ^{2} x d x=\int\left(\frac{\sin 2 x}{2}\right)^{2} \times \\
\times\left(\frac{1-\cos 2 x}{2}\right) d x=\frac{1}{8} \int \sin ^{2} 2 x(1-\cos 2 x) d x=\frac{1}{8} \int \sin ^{2} 2 x d x- \\
-\frac{1}{8} \int \sin ^{2} 2 x \cos 2 x d x=\frac{1}{8} \int \frac{1-\cos 4 x}{2} d x-\frac{1}{8} \int \sin ^{2} 2 x \times \\
\times \frac{1}{2} d(\sin 2 x)=\frac{1}{16} \int d x-\frac{1}{16} \int \cos 4 x d x- \\
-\frac{1}{16} \int \sin ^{2} 2 x d(\sin 2 x)=\frac{1}{16} x-\frac{1}{64} \sin 4 x-\frac{1}{48} \sin ^{3} 2 x+C
\end{gathered}
$$