Example 3. Find the integral $\int \sin ^{4} x d x$.
Solution. Here $n=4, m=0$. Transforming the integrand using the appropriate formulas, we find

$$
\begin{gathered}
\int \sin ^{4} x d x=\int\left(\sin ^{2} x\right)^{2} d x=\int\left(\frac{1-\cos 2 x}{2}\right)^{2} d x= \\
=\frac{1}{4} \int\left(1-2 \cos 2 x+\cos ^{2} 2 x\right) d x=\frac{1}{4} \int d x-\frac{1}{4} \int \cos 2 x d(2 x)+ \\
+\frac{1}{4} \int \frac{1}{2}(1+\cos 4 x) d x=\frac{1}{4} \int d x-\frac{1}{4} \int \cos 2 x d(2 x)+ \\
+\frac{1}{8} \int d x+\frac{1}{8 \cdot 4} \int \cos 4 x d(4 x)=\frac{1}{4} x-\frac{1}{4} \sin 2 x+ \\
+\frac{1}{8} x+\frac{1}{32} \sin 4 x+C=\frac{3}{8} x-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C
\end{gathered}
$$