<s>[INST] Integral of $\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}$dx Question:
$$\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}\mathrm dx.$$
What we did:
we tried using $t=\tan (\frac x2)$ and also dividing both numerator and denominator by $\sqrt {\cos x}$, eventually using the second method we got to this:
$\displaystyle \int \frac {2t+2}{t^2+2t-1}-\frac {2}{t^2+2t-1} +\frac {\sqrt{2t(1-t^2)}}{t^2+2t-1}  $, for which we know how to solve the first and second integral but not the third...
Thanks
 [/INST] Alternatively, if we call our integral $I$ and perform a dummy variable substitution, it all falls out rather nicely! $$I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x \ \overset{x \mapsto \frac{\pi}{2} - x}= \ \int_{0}^{\pi/2} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x $$ Adding the two integrals together gives $$ 2I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x = \int_{0}^{\pi/2} \text{d}x = \dfrac{\pi}{2} \implies \ I = \dfrac{\pi}{4}$$
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