Answer
$$\ln \left( 2 \right) - \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {{{\tan }^5}\frac{x}{2}} dx \cr
& {\text{Write the integrand as}} \cr
& = \int_0^{\pi /2} {{{\tan }^3}\left( {\frac{x}{2}} \right)} {\tan ^2}\left( {\frac{x}{2}} \right)dx \cr
& {\text{Use the trigonometric identity ta}}{{\text{n}}^2}\theta + 1 = {\sec ^2}\theta \cr
& = \int_0^{\pi /2} {{{\tan }^3}\left( {\frac{x}{2}} \right)} \left( {{{\sec }^2}\left( {\frac{x}{2}} \right) - 1} \right)dx \cr
& {\text{Multiply}} \cr
& = \int_0^{\pi /2} {\left( {{{\tan }^3}\left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right) - {{\tan }^3}\left( {\frac{x}{2}} \right)} \right)} dx \cr
& = \int_0^{\pi /2} {\left( {{{\tan }^3}\left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right)\left( {{{\sec }^2}\left( {\frac{x}{2}} \right) - 1} \right)} \right)} dx \cr
& = \int_0^{\pi /2} {\left( {{{\tan }^3}\left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right) + \tan \left( {\frac{x}{2}} \right)} \right)} dx \cr
& = 2\int_0^{\pi /2} {\left[ {{{\tan }^3}\left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right)\left( {\frac{1}{2}} \right) - \tan \left( {\frac{x}{2}} \right){{\sec }^2}\left( {\frac{x}{2}} \right)\left( {\frac{1}{2}} \right) + \tan \left( {\frac{x}{2}} \right)\left( {\frac{1}{2}} \right)} \right]} dx \cr
& {\text{Integrate}} \cr
& = 2\left( {\frac{1}{4}{{\tan }^4}\left( {\frac{x}{2}} \right) - \frac{1}{2}\tan \left( {\frac{x}{2}} \right) - \ln \left| {\cos \frac{x}{2}} \right|} \right)_0^{\pi /2} \cr
& = \left( {\frac{1}{2}{{\tan }^4}\left( {\frac{\pi }{4}} \right) - \tan \left( {\frac{\pi }{4}} \right) - 2\ln \left| {\cos \frac{\pi }{4}} \right|} \right) - \left( {\frac{2}{3}{{\tan }^3}\left( 0 \right) - \tan \left( 0 \right) - 2\ln \left| {\cos 0} \right|} \right) \cr
& {\text{Simplifying}} \cr
& = \left( {\frac{1}{2} - 1 - 2\ln \left| {\frac{{\sqrt 2 }}{2}} \right|} \right) - \left( 0 \right) \cr
& = - \frac{1}{2} - 2\ln \left( {\frac{{\sqrt 2 }}{2}} \right) \cr
& = - \frac{1}{2} - \ln \left( {\frac{2}{4}} \right) \cr
& = \ln \left( 2 \right) - \frac{1}{2} \cr} $$
