Answer
$\frac{2}{3}tan^{3}\frac{x}{2}+2tan\frac{x}{2}+C$
Work Step by Step
$\int\sec^{4}(\frac{x}{2})dx$
$\int\sec^{2}\frac{x}{2}sec^{2}(\frac{x}{2})dx$
By the pythagorean identity $tan^{2}x+1=sec^{2}x$:
$\int(tan^{2}\frac{x}{2}+1)sec^{2}(\frac{x}{2})dx$
$let$ $u=tan\frac{x}{2}$
$du=\frac{1}{2}sec^{2}\frac{x}{2}$
$2du=sec^{2}\frac{x}{2}$
$2\int(u^{2}+1)du$
$2[\frac{1}{3}u^{3}+u]+C$
$2[\frac{1}{3}tan^{3}\frac{x}{2}+tan\frac{x}{2}]+C$
$\frac{2}{3}tan^{3}\frac{x}{2}+2tan\frac{x}{2}+C$
