Example 1.11. Find $\int \frac{\sin ^{3} x}{\cos ^{5} x} d x$.
Solution. Since the integrand is odd with respect to $\sin x$, applying the substitution $t=\cos x$ we get, considering that $\sin ^{2} x=1-\cos ^{2} x, d t=-\sin x d x$ :

$$
\begin{gathered}
\int \frac{\sin ^{3} x}{\cos ^{5} x} d x=\int \frac{1-\cos ^{2} x}{\cos ^{5} x} \sin x d x=-\int \frac{1-t^{2}}{t^{5}}= \\
=-\int t^{-5} d t+\int t^{-3} d t=\frac{1}{4 t^{4}}-\frac{1}{2 t^{2}}+C= \\
=\frac{1}{4 \cos ^{4} x}-\frac{1}{2 \cos ^{2} x}+C
\end{gathered}
$$

Integrals of the form $\int \sin ^{2 m} x \cos ^{2 n} x d x$, where $m$ and $n$ are natural numbers, can be found using the substitution $t=\operatorname{tg} x$. However, transforming the integrand using the formulas

$$
\sin x \cos x=\frac{1}{2} \sin 2 x, \sin ^{2} x=\frac{1}{2}(1-\cos 2 x), \cos ^{2} x=\frac{1}{2}(1+\cos 2 x)
$$

is much faster.