Example 10. Find the integral $\int \frac{d x}{\sqrt{\left(x^{2}-a^{2}\right)^{3}}}$.
Solution. Apply the trigonometric substitution

$$
x=a \sec t
$$

We have:

$$
\begin{gathered}
d x=\frac{a \sin t d t}{\cos ^{2} t}=\frac{a \operatorname{tg} t d t}{\cos t} \\
\sqrt{\left(x^{2}-a^{2}\right)^{3}}=\sqrt{\left(a^{2} \sec ^{2} t-a^{2}\right)^{3}}=\sqrt{a^{6}\left(\frac{1-\cos ^{2} t}{\cos ^{2} t}\right)^{3}}= \\
=\sqrt{a^{6} \operatorname{tg}^{6} t}=a^{3} \operatorname{tg}^{3} t \\
\int \frac{d x}{\sqrt{\left(x^{2}-a^{2}\right)^{3}}}=\int \frac{1}{a^{3} \operatorname{tg}^{3} t} \cdot \frac{a \operatorname{tg} t d t}{\cos t}=\frac{1}{a^{2}} \int \frac{d t}{\operatorname{tg}^{2} t \cos t}= \\
=\frac{1}{a^{2}} \int \frac{\cos t d t}{\sin ^{2} t}=\frac{1}{a^{2}} \int(\sin t)^{-2} d(\sin t)=\frac{1}{a^{2}} \frac{(\sin t)^{-1}}{-1}+C= \\
=-\frac{1}{a^{2}} \cdot \frac{1}{\sin t}+C=-\frac{1}{a^{2} \sin t}+C
\end{gathered}
$$

Returning to the variable $x$. Since

$$
x=a \sec t
$$

then

$$
\sec t=\frac{x}{a}, \frac{1}{\cos t}=\frac{x}{a}, \cos t=\frac{a}{x}
$$

We find $\sin t$ :

$$
\sin t=\sqrt{1-\cos ^{2} t}=\sqrt{1-\frac{a^{2}}{x^{2}}}=\sqrt{\frac{x^{2}-a^{2}}{x^{2}}}=\frac{\sqrt{x^{2}-a^{2}}}{x}
$$

Therefore,

$$
\int \frac{d x}{\sqrt{\left(x^{2}-a^{2}\right)^{3}}}=-\frac{1}{a^{2} \sin t}+C=-\frac{1}{a^{2}} \frac{x}{\sqrt{x^{2}-a^{2}}}+C
$$