Solution:

Assume $I=\int \tan ^{5} x \sec ^{4} x d x$
We can write the above equation as
$\Rightarrow I=\int \tan ^{5} x \sec ^{2} x \sec ^{2} x d x$
Taking $\tan ^{5} \mathrm{x}$ as common, we get
$\begin{array}{l}
\Rightarrow I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \sec ^{2} x d x \\
\Rightarrow I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \sec ^{2} x d x
\end{array}$
On simplifying, we get
$\Rightarrow I=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$
Suppose $\tan x=t$, therefore
$\Rightarrow \sec ^{2} x d x=d t$
On substituting the value of $x$, we get
$\Rightarrow I=\int\left(t^{5}+t^{7}\right) d t$
On integrating we get
$\Rightarrow \mathrm{I}=\frac{\mathrm{t}^{6}}{6}+\frac{\mathrm{t}^{8}}{8}+\mathrm{c}$
Substituting the values of $t$
$\Rightarrow \mathrm{I}=\frac{\tan ^{6} \mathrm{x}}{6}+\frac{\tan ^{8} \mathrm{x}}{8}+\mathrm{c}$
Hence, $\int \tan ^{5} x \sec ^{4} x d x=\frac{\tan ^{6} x}{6}+\frac{\tan ^{8} x}{8}+c$