The problem: Find the value of $\displaystyle\int_0^{\pi/2} \frac{dx}{1+(\tan(x))^{\sqrt{2}}}$
I tried a few different substitutions and the closest I got to an okay looking integral is $$\int_0^\infty \frac{t^2 \, dt}{t^{1/\sqrt{2}}(1+\sqrt{t})},$$ which still looks scary. How do I approach this integral?
It actually works for $$ I=\int_0^{\pi/2}\frac{dx}{1+(\tan x)^r},\ \ \ r\geq0 $$
Let $y=\pi/2-x$. Then $$ I=\int_{\pi/2}^0\frac{-dy}{1+(\tan(\pi/2-x))^r}=\int_0^{\pi/2}\frac{dy}{1+\left(\frac1{\tan y}\right)^r} =\int_0^{\pi/2}\frac{(\tan y)^r}{1+(\tan y)^r}\,dy. $$ Then $$ 2I=I+I=\int_0^{\pi/2}\frac{dx}{1+(\tan x)^r}+\int_0^{\pi/2}\frac{(\tan x)^r}{1+(\tan x)^r}\,dx=\int_0^{\pi/2}\frac{1+(\tan x)^r}{1+(\tan x)^r}\,dx=\frac\pi2. $$ Then $$ I=\frac\pi4. $$
Hint. One may perform the change of variable $u=\dfrac \pi2 -x$ and observe that $$ \left(\tan \left(\frac \pi2 -u\right)\right)^{\sqrt{2}}=\frac1{\left(\tan u\right)^{\sqrt{2}}} $$ giving $$ \int_0^{\pi/2}\frac{dx}{1+\left(\tan x\right)^{\sqrt{2}}}=\int_0^{\pi/2}\frac{\left(\tan x\right)^{\sqrt{2}}dx}{1+\left(\tan x\right)^{\sqrt{2}}}. $$
