Answer
$$\frac{1}{2}{\tan ^4}\frac{x}{2} - {\tan ^2}\frac{x}{2} - 2\ln \left| {\cos \frac{x}{2}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^5}\frac{x}{2}} dx \cr
& {\text{Split the integrand, use }}{a^{m + n}} = {a^m}{a^n} \cr
& = \int {{{\tan }^3}\frac{x}{2}{{\tan }^2}\frac{x}{2}} dx \cr
& {\text{Use the identity }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& = \int {{{\tan }^3}\frac{x}{2}\left( {{{\sec }^2}\frac{x}{2} - 1} \right)} dx \cr
& = \int {{{\tan }^3}\frac{x}{2}{{\sec }^2}\frac{x}{2}} dx - \int {{{\tan }^3}\frac{x}{2}dx} \cr
& = \int {{{\tan }^3}\frac{x}{2}{{\sec }^2}\frac{x}{2}} dx - \int {\tan \frac{x}{2}{{\tan }^2}\frac{x}{2}dx} \cr
& = \int {{{\tan }^3}\frac{x}{2}{{\sec }^2}\frac{x}{2}} dx - \int {\tan \frac{x}{2}\left( {{{\sec }^2}\frac{x}{2} - 1} \right)dx} \cr
& = \int {{{\tan }^3}\frac{x}{2}{{\sec }^2}\frac{x}{2}} dx - \int {\tan \frac{x}{2}{{\sec }^2}\frac{x}{2}dx} + \int {\tan \frac{x}{2}dx} \cr
& {\text{Rewrite integrands}} \cr
& 2\int {{{\tan }^3}\frac{x}{2}{{\sec }^2}\frac{x}{2}\left( {\frac{1}{2}} \right)} dx - 2\int {\tan \frac{x}{2}{{\sec }^2}\frac{x}{2}\left( {\frac{1}{2}} \right)dx} + 2\int {\tan \frac{x}{2}\left( {\frac{1}{2}} \right)dx} \cr
& {\text{Integrate}} \cr
& = 2\left( {\frac{{{{\tan }^4}\frac{x}{2}}}{4}} \right) - 2\left( {\frac{{{{\tan }^2}\frac{x}{2}}}{2}} \right) - 2\ln \left| {\cos \frac{x}{2}} \right| + C \cr
& = \frac{1}{2}{\tan ^4}\frac{x}{2} - {\tan ^2}\frac{x}{2} - 2\ln \left| {\cos \frac{x}{2}} \right| + C \cr} $$
