<s>[INST] Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$? 
Which is the easiest way to evaluate $\int \limits_{0}^{\pi/2} (\sqrt{\tan x} +\sqrt{\cot x})$?

I have reduced this problem to $$   2\int_0^{\pi/2} \sqrt{\tan x} \ dx$$
but now, evaluating this integral is giving me some problems, simply substituting $u=\tan(x)$
and then $\mathrm{d}u=\sec^2(x)\mathrm{d}x \Rightarrow \frac{\mathrm{d}u}{1+u^2}=\mathrm{d}x$ and which in turn gives something a bit ugly, I was wondering which is the most elegant way to evaluate this?
 [/INST] $${\int_0^{\frac{\pi}{2}} \sqrt{\tan x}dx + \sqrt{\cot x}dx}$$
$$={\int_0^{\frac{\pi}{2}}\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}dx = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\frac{\sqrt{2\sin{x}\cos{x}}}{\sqrt{2}}}dx = \sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (1 - 2 \sin{x} \cos{x})}}dx}$$
$${=\sqrt{2}\int_0^{\frac{\pi}{2}} \frac{ \sin{x} + \cos{x}}{\sqrt{1 - (\sin{x} - \cos{x})^2}}dx}$$
Let ${t = \sin{x} - \cos{x}}$, $\Large {{\small{dx}} = \frac{dt}{\sin{x} + \cos{x}}}$ $${x \to \frac{\pi}{2} \implies t = (\sin{x} - \cos{x}) \to 1}$$ $${x \to 0 \implies t = (\sin{x} - \cos{x}) \to -1}$$
$$\sqrt{2}\int_{-1}^{1} \frac{1}{\sqrt{1 - t^2}}dt = \sqrt{2}\left[\sin^{-1}{t}\right]_{-1}^{1} = \sqrt{2}\left[\frac{\pi}{2} - \left(- \frac{\pi}{2} \right) \right] = \sqrt{2} \pi $$

I think this might be the simplest approach.
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