Example 7 (to $5^{\circ}$ ). Find $\int \frac{\sin ^{5} 3 x}{\cos ^{5} 3 x} d x$.
Solution. In this case, it is easier to switch to the tangent. We will show all transformations in the course of the solution. We have:

$$
\begin{aligned}
& \int \frac{\sin ^{5} 3 x}{\cos ^{5} 3 x} d x=\int \operatorname{tg}^{5} 3 x d x=\int \operatorname{tg}^{3} 3 x\left(\frac{1}{\cos ^{2} 3 x}-1\right) d x= \\
& =\int \operatorname{tg}^{3} 3 x \cdot \frac{d x}{\cos ^{2} 3 x}-\int \operatorname{tg}^{3} 3 x d x=\frac{1}{3} \int \operatorname{tg}^{3} 3 x d(\operatorname{tg} 3 x)- \\
& -\int \operatorname{tg} 3 x\left(\frac{1}{\cos ^{2} 3 x}-1\right) d x=\frac{\operatorname{tg}^{4} 3 x}{12}-\frac{1}{3} \int \operatorname{tg} 3 x d(\operatorname{tg} 3 x)+\int \frac{\sin 3 x}{\cos 3 x} d x= \\
& =\frac{\operatorname{tg}^{4} 3 x}{12}-\frac{1}{6} \operatorname{tg}^{2} 3 x-\frac{1}{3} \ln |\cos 3 x|+C
\end{aligned}
$$

In some examples, you can save time by using trigonometric transformations.