Compute $ \int_{0}^{\pi/2}\tan^n x \ dx$ Compute $$ \int_{0}^{\pi/2}\tan^n x \ dx$$
Using reduction formula I got $$I_n = \int_0^{\pi/2} \tan^n(x) dx = \int_0^{\pi/2} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/2} \tan^{n-2}(x)dx$$
But I'm not sure how to proceed... 
 A: $$
\int_0^{\frac{\pi}{2}}\tan^nxdx= \int_0^{\frac{\pi}{4}}\tan^nxdx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan^nxdx > \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan x dx = \ln\left(\frac{\sec(\pi/2)}{\sec(\pi/4)}\right) = \infty
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\tan^{n}\pars{x} & 
\,\,\,\stackrel{\mrm{as}\ x\ \to\ 0^{+}}{\sim}\,\,\, x^{n}
\\[5mm]
\tan^{n}\pars{x} & = \cot^{n}\pars{{\pi \over 2} - x} = {1 \over \tan^{n}\pars{\pi/2 - x}}
\,\,\,\stackrel{\mrm{as}\ x\ \to\ \pars{\pi/2}^{-}}{\sim}\,\,\,
\pars{{\pi \over 2} - x}^{-n}
\end{align}

The integral converges whenever $\ds{\verts{\Re\pars{n}} < 1}$.

