Example 1.12. Find $\int \sin ^{4} x \cos ^{4} x d x$.
Solution.

$$
\begin{gathered}
\int \sin ^{4} x \cos ^{4} x d x=\frac{1}{16} \int(2 \sin x \cos x)^{4} d x= \\
=\frac{1}{16} \int \sin ^{4} 2 x d x=\frac{1}{16} \int\left(\frac{1-\cos 4 x}{2}\right)^{2} d x= \\
=\frac{1}{64} \int\left(1-2 \cos 4 x+\cos ^{2} 4 x\right) d x= \\
=\frac{1}{64}\left(\int d x-2 \int \cos 4 x d x+\frac{1}{2} \int(1+\cos 8 x) d x\right)= \\
=\frac{1}{64}\left(x-\frac{1}{2} \int \cos 4 x d 4 x+\frac{1}{2} \int d x+\frac{1}{16} \int \cos 8 x d(8 x)\right)= \\
=\frac{1}{64}\left(\frac{3}{2} x-\frac{1}{2} \sin 4 x+\frac{1}{16} \sin 8 x\right)+C= \\
=\frac{1}{128}\left(3 x-4 \sin 4 x+\frac{1}{8} \sin 8 x\right)+C
\end{gathered}
$$

Integrals of the form $\int \sin \alpha x \cos \beta x d x, \int \cos \alpha x \cos \beta x d x$, $\int \sin \alpha x \sin \beta x d x$ (where $\alpha$ and $\beta$ are constants) are found by transforming the integrand into a sum of trigonometric functions using the formulas:

$$
\begin{aligned}
& \sin \alpha x \cos \beta x=\frac{1}{2}(\sin (\alpha+\beta) x+\sin (\alpha-\beta) x) \\
& \cos \alpha x \cos \beta x=\frac{1}{2}(\cos (\alpha+\beta) x+\cos (\alpha-\beta) x) \\
& \sin \alpha x \sin \beta x=\frac{1}{2}(\cos (\alpha-\beta) x-\cos (\alpha+\beta) x)
\end{aligned}
$$