Calcular $ \int_{0}^{\pi/2}\tan^n x \ dx$

Question

Calcular $$ \int_{0}^{\pi/2}\tan^n x \ dx$$

Utilizando la fórmula de reducción conseguí $$I_n = \int_0^{\pi/2} \tan^n(x) dx = \int_0^{\pi/2} \tan^{n-2}(x) \sec^2(x) dx - \int_0^{\pi/2} \tan^{n-2}(x)dx$$ Pero no estoy seguro de cómo proceder...

Answer 1

$$ \int_0^{\frac{\pi}{2}}\bronceado^nxdx= \int_0^{\frac{\pi}{4}}\bronceado^nxdx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\bronceado^nxdx > \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\tan x dx = \ln\left(\frac{\s(\pi/2)}{\s(\pi/4)}\right) = \infty $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

La integral converge siempre que $\ds{\verts{\Re\pars{n}} < 1}$.