Example 4 (to $4^{\circ}$ ). Find $\int \sin ^{2} x \cos ^{4} x d x$.
Solution. We are dealing with case 2). We reduce the degree of the integrand expression twice, then transform the product into a sum and combine like terms:

$$
\begin{gathered}
\int \sin ^{2} x \cos ^{2} x \cdot \cos ^{2} x d x=\int \frac{1}{4} \sin ^{2} 2 x \cdot \frac{1}{2}(1+\cos 2 x) d x= \\
=\frac{1}{8} \int \frac{1}{2}(1-\cos 4 x) \cdot(1+\cos 2 x) d x= \\
=\frac{1}{16} \int(1-\cos 4 x \cos 2 x-\cos 4 x+\cos 2 x) d x= \\
=\frac{1}{16} \int\left(1-\frac{1}{2}(\cos 6 x+\cos 2 x)-\cos 4 x+\cos 2 x\right)= \\
=\frac{1}{16}\left(x-\frac{1}{12} \sin 6 x+\frac{1}{4} \sin 2 x-\frac{1}{4} \sin 4 x\right)
\end{gathered}
$$

Note. We could have proceeded differently, starting from the second equality sign (omitting the coefficient $\frac{1}{8}$):

$$
\begin{gathered}
\int \sin ^{2} 2 x(1+\cos 2 x) d x=\int \sin ^{2} 2 x d x+\int \sin ^{2} 2 x \cos 2 x d x= \\
=\int \frac{1-\cos 4 x}{2} d x+\frac{1}{2} \int \sin ^{2} 2 x d(\sin 2 x)= \\
=\frac{1}{2} x-\frac{1}{8} \sin 4 x+\frac{1}{6} \sin ^{3} 2 x+C
\end{gathered}
$$