Definite integral of $\frac{1}{\sqrt{\tan x}}$ i would like to ask you for help to evaluate  $$\int_{0}^{\pi/2}\frac{1}{\sqrt{\tan x}}dx$$ 
I didn't find an appropriate substitution.
Thanks for help.
 A: Hint
As in the previous answer, let
$$z=\sqrt{\tan(x)}\implies x=\tan ^{-1}\left(z^2\right)\implies dx=\frac{2 z}{z^4+1}\,dz$$ This makes $$\int\frac{dx}{\sqrt{\tan x}}=2\int \frac{dz}{z^4+1}$$ Now $$z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt 2 z+1)(z^2-\sqrt 2 z+1)$$ Now, using partial fraction decomposition $$\frac 1{z^4+1}=\frac 1{(z^2+\sqrt 2 z+1)(z^2-\sqrt 2 z+1)}$$ $$\frac 1{z^4+1}=\frac 1{2\sqrt 2}\left(\frac{z+\sqrt{2}}{z^2+\sqrt{2} z+1}-\frac{z-\sqrt{2}}{z^2-\sqrt{2} z+1}\right)$$ which seems to lead to quite standard integrations leading to some $\tan^{-1}(.)$ and $\log(.)$.
A: $$I = \int\limits_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{\tan x}} dx$$
$$z = \sqrt{\tan x}$$
$$dz = \frac{\sec^2x}{2\sqrt{\tan x}}dx$$
Therefore,
$$I = \int\limits_{0}^{\infty}\frac{2}{\sec^2x}dz = \int\limits_{0}^{\infty}\frac{2}{z^4+1}dz \,\,\,\,(\sec^2x = 1+\tan^2x)$$
Use partial fraction from here. Please have a look at @Claude answer.
A: If one does not want to proceed by partial fraction decomposition, one can do a trick: 
First, $z=\sqrt{\tan x}$ will give you, as mentioned in other answers,
$$
\int_0^{\pi/2}\frac{1}{\sqrt{\tan x}}\,dx=\int_0^{+\infty}\frac{2}{1+z^4}\,dz=\int_0^1\frac{2}{1+z^4}\,dz
+\int_1^{+\infty}\frac{2}{1+z^4}\,dz
$$
In the second integral, do $z\mapsto 1/z$, and you will get that
$$
\int_0^{\pi/2}\frac{1}{\sqrt{\tan x}}\,dx
=
\int_0^1\frac{2+2z^2}{1+z^4}\,dz.
$$
Thus,
$$
\begin{aligned}
\int_0^{\pi/2}\frac{1}{\sqrt{\tan x}}\,dx&=
\int_0^1\frac{2+2z^2}{1+z^4}\,dz\\
&=\int_0^1\frac{2/z^2+2}{1/z^2+z^2}\,dz\\
&=\int_0^1\frac{(1+1/z^2)}{1+\bigl((z-1/z)/\sqrt{2}\bigl)^2}\,dz\\
&=\Bigl[\sqrt{2}\arctan\bigl((z-1/z)/\sqrt{2}\bigr)\Bigr]_0^1\\
&=\frac{\pi}{\sqrt{2}}.
\end{aligned}
$$
A: We may also use the residue theorem to compute the integral. Note that $$I=\int_{0}^{\infty}\frac{2}{1+z^{4}}dz=\int_{-\infty}^{\infty}\frac{1}{1+z^{4}}dz$$ and now take the upper half circle with radius $R$ as contour. It is not difficult to see that the inegral over the semicircumference vanish as $R\rightarrow\infty$ so $$\int_{-\infty}^{\infty}\frac{1}{1+z^{4}}dz=2\pi i\left(\underset{z=e^{i\pi/4}}{\textrm{Res}}\frac{1}{1+z^{4}}+\underset{z=e^{i3\pi/4}}{\textrm{Res}}\frac{1}{1+z^{4}}\right)=\color{red}{\frac{\pi}{\sqrt{2}}}.$$ 
