<s>[INST] 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.
 [/INST] 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 $\log(.)$.
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