Solution:

$\mathrm{I}=\int \frac{x^{7}}{\left(a^{2}-x^{2}\right)^{5}} d x$
Suppose $\mathrm{x}=a \sin \theta$
On differentiating both the sides we obtain
$d x=a \cos \theta d \theta$
$\begin{array}{l}
\therefore I=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{\left(a^{2}-a^{2} \sin ^{2} \theta\right)^{5}} d \theta \\
=\int \frac{a^{8} \sin ^{7} \theta \cos \theta}{a^{10}\left(1-\sin ^{2} \theta\right)^{5}} d \theta \\
=\int \frac{\sin ^{7} \theta}{a^{2} \cos ^{9} \theta} d \theta \\
=\frac{1}{a^{2}} \int \tan ^{7} \theta \sec ^{2} \theta d \theta
\end{array}$
Suppose
$\tan \theta=t$
Again on differentiating on both the sides, we get
$\begin{array}{l}
\sec ^{2} \theta d \theta=d t \\
\therefore I=\frac{1}{a^{2}} \int t^{7} d t \\
=\frac{1}{a^{2}} \frac{t^{8}}{8}+c
\end{array}$
$\begin{array}{l}
=\frac{1}{8 a^{2}}\left(\tan ^{8} \theta\right)+c \\
=\frac{1}{8 a^{2}}\left(\tan \left(\sin ^{-1} \frac{x}{a}\right)\right)^{8}+c \\
=\frac{1}{8 a^{2}}\left(\tan \left(\tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}\right)\right)^{8}+c
\end{array}$
$=\frac{1}{8 a^{2}}\left(\frac{x}{\sqrt{a^{2}-x^{2}}}\right)^{8}+c$
$=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{4}}+c$
As a result, $\int \frac{x^{7}}{\left(a^{2}-x^{2}\right)^{5}} d x=\frac{1}{8 a^{2}} \frac{x^{8}}{\left(a^{2}-x^{2}\right)^{4}}+c$