Answer
$\frac{1}{20}\tan^4 5x+C$
Work Step by Step
Let $u=\tan 5x$. Then $du=5\sec^2 5xdx$, so $\sec^2 5xdx=\frac{1}{5}du$.
$\int \tan^3 5x\sec^2 5xdx$
$=\int u^3*\frac{1}{5}du$
$=\frac{1}{5}*\frac{u^4}{4}+C$
$=\frac{1}{20}u^4+C$
$=\frac{1}{20}\tan^4 5x+C$
