Different Solution for $\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\sqrt{2}}}$ I Believe that I have the optimal solution, using the fact that: 
$$\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\sqrt{2}}}=\int_{0}^{\pi/2}\frac{\cos(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}=\int_{0}^{\pi/2}\frac{\sin(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}$$
and the noticing that 
$$\int_{0}^{\pi/2}\frac{\cos(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}+\int_{0}^{\pi/2}\frac{\sin(x)^{\sqrt{2}}}{\sin(x)^{\sqrt{2}}+\cos(x)^{\sqrt{2}}}=\pi/2.$$
These together imply that the solution is $\pi/4$.
remark: the $\sqrt{2}$ is unimportant, for the more general $\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\alpha}}$,  my solution is invariant for any  $\alpha \in {\mathbb{R}^{+}}$.
However, had I been given 
$$\int_{0}^{\pi/2}\frac{1}{1+\tan(x)}$$
I maybe would have actually attempted a solution by more standard methods, in order to find a closed expression for the integral. My method does not give a closed form, does anyone know a way to find one?
I was given a possible alternative route:
Let $$F(\alpha)=\int_{0}^{\pi/2}\frac{1}{1+\tan(x)^{\alpha}},$$
and to differentiate with respect to $\alpha$, yet I'm not sure why this is helpful or how that kind of method "goes." 
 A: Differentiation under the integral sign is another viable alternative, but the trick is pretty the same:
$$ \frac{dF}{d\alpha}=\int_{0}^{\pi/2}\left(-\log\tan x\right)\frac{\tan(x)^\alpha}{(1+\tan(x)^\alpha)^2}\,dx\\=-\frac{1}{2}\int_{0}^{\pi/2}\left(\log\tan x+\log\cot x\right)\frac{\tan(x)^\alpha}{(1+\tan(x)^\alpha)^2}\,dx =\color{red}{0}$$
gives that $F(\alpha)$ is constant over $\mathbb{R}^+$, and
$$ \lim_{\alpha\to 0^+} F(\alpha) = \int_{0}^{\pi/2}\frac{dx}{2}=\color{red}{\frac{\pi}{4}}.$$

An alternative proof may go through the following lines. We have:
$$ F(\alpha) = \int_{0}^{+\infty}\frac{du}{(1+u^2)(1+u^\alpha)}.$$
Through the substitution $u=v^{-1}$ we have:
$$ F(\alpha)=\int_{0}^{+\infty}\frac{u^{\alpha}\,du}{(1+u^2)(1+u^{\alpha})}$$
so:
$$ F(\alpha)=\frac{1}{2}\int_{0}^{+\infty}\frac{du}{1+u^2}=\frac{\pi}{4}.$$
A: Let $$I= \int_{0}^{\pi/2}\frac{dx}{1+\tan x}$$
You can  replace $x \rightarrow \pi/2 -x$ in the integral to get
$$I=\int_{0}^{\pi/2}\frac{dx}{1+\cot x}$$
$$\implies I=\int_{0}^{\pi/2}\frac{\tan xdx}{1+\tan x}$$
Adding this to the original integral gives 
$$I=\frac{\pi}{4}$$
