show that $\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$ show that $$\int_{0}^{\pi/2}\tan^ax \, dx=\frac {\pi}{2\cos(\frac{\pi a}{2})}$$
I think we can solve it by contour integration but I dont know how. 
If someone can solve it by two way using complex and real analysis its better for me. 
thanks for all.

Sorry for being late
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \tan ^a x d x&=\int_0^{\frac{\pi}{2}} \sin ^a x \cos ^{-a} x d x \\
& =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} x \cos ^{2\left(\frac{-a+1}{2}\right)-1} x d x \\
& =\frac{1}{2} B\left(\frac{a+1}{2}, \frac{-a+1}{2}\right) \\
& =\frac{1}{2} \pi \csc \frac{(a+1) \pi}{2} \\
&
\end{aligned}
$$
Applying the Euler-reflection property
$$
B(x, 1-x)=\pi \csc (\pi x) \quad x \notin \mathbb{Z},
$$
we have
$$\boxed{\int_0^{\frac{\pi}{2}} \tan ^a x d x =\frac{\pi}{2 \cos \left(\frac{\pi a}{2}\right)}} $$
