intergral including tan x

[Math]

$

\displaystyle \int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}$

Could anyone tell me how to start,

In a book ,it was given that $\sqrt2$ does not even matter

 

[Markov]

Replace $\sqrt2$ for any power and put $x\mapsto\dfrac\pi2-x.$

 

[sbhatnagar]

\( \displaystyle \begin{align*}
I &=\int_{0}^{\pi/2}\frac{1}{1+\tan^{\sqrt{2}}(x)}dx \\ \Rightarrow I &=\int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (1)

\end{align*} \)

Applying Property,


\( \displaystyle \begin{align*}
I &= \int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}{\sin^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)+ \cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}dx\\ \Rightarrow I &= \int_{0}^{\pi/2}\frac{\sin^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (2)

\end{align*} \)

Add equations (1) and (2) and see what happens!

 

[Sherlock]

Let $x = \tan^{-1}{t}$, then:

$\begin{aligned} I & = \int_{0}^{\pi/2}\frac{1}{1+\tan^{\alpha}{x}}\;{dx} \\& = \int_{0}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt} \end{aligned}$

Range it from $[0, 1]$ and $[1, \infty]$,

$\displaystyle I = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{1}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}$

Put $t \mapsto \frac{1}{t}$ for the second one,

$\displaystyle \begin{aligned}I & = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{0}^{1}\frac{t^{\alpha}}{(1+t^2(1+t^{\alpha})}\;{dt} \\& = \int_{0}^{1}\frac{1+t^{\alpha}}{(1+t^2)(1+t^{ \alpha})}\;{dt} = \int_{0}^{1}\frac{1}{1+t^2}\;{dt} = \frac{\pi}{4}.\end{aligned}$