Answer
$\dfrac{2\tan^3\left(\frac{x}{2}\right)}{3}+2\tan\left(\dfrac{x}{2}\right)+C$
Work Step by Step
$I= \int sec^4(\frac{x}{2})dx = \int sec^2(\frac{x}{2})sec^2(\frac{x}{2})dx$
Using $sec^2(\theta)=tan^2(\theta)+1$ We get:
$\int sec^2(\frac{x}{2})[tan^2(\frac{x}{2})+1]dx = \int sec^2(\frac{x}{2})tan^2(\frac{x}{2})dx+\int sec^2(\frac{x}{2})dx$
$I_1=\int sec^2(\frac{x}{2})tan^2(\frac{x}{2})dx$
$I_2=\int sec^2(\frac{x}{2})dx$
$I_2= 2tan(\frac{x}{2})+C_2$
Using $\int f'(x)f^n(x) dx = \frac{f^{n+1}}{n+1}+C$ for $I_1$ We get:
$I_1=2\int \frac{1}{2}sec^2(\frac{x}{2})tan^2(\frac{x}{2})dx =\frac{2tan^3(\frac{x}{2})}{3}+C_1$
$I=I_1+I_2=\dfrac{2\tan^3\left(\frac{x}{2}\right)}{3}+2\tan\left(\dfrac{x}{2}\right)+C.$
