Answer
$$\int\tan^7\frac{x}{2}\sec^2\frac{x}{2}dx=\frac{1}{4}\tan^8\frac{x}{2}+C$$
Work Step by Step
$$A=\int\tan^7\frac{x}{2}\sec^2\frac{x}{2}dx$$
We set $u=\tan\frac{x}{2}$
Then $$du=\frac{1}{2}\sec^2\frac{x}{2}dx$$
That means $$\sec^2\frac{x}{2}dx=2du$$
Therefore, $$A=2\int u^7du=\frac{2u^8}{8}+C=\frac{u^8}{4}+C$$ $$A=\frac{1}{4}\tan^8\frac{x}{2}+C$$