Integral of $\int_0^{\pi/2} \frac {\sqrt{\sin x}}{\sqrt {\sin x} + \sqrt {\cos x}}$dx [duplicate]

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

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 $$ 2I = \int_{0}^{\pi/2} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \text{ d}x = \int_{0}^{\pi/2} \text{d}x = \dfrac{\pi}{2} \implies \ I = \dfrac{\pi}{4}$$

You mentioned dividing numerator and denominator by $\sqrt{\cos x}$.

$$\int\frac{\sqrt{\tan x}}{\sqrt{\tan x}+1}dx$$

Did you consider the substitution $x=\tan^{-1}u^2$?

$$dx=\frac{2udu}{1+u^4}$$

$$\int\frac{2u^2du}{(u+1)(u^4+1)}$$

I don't remember offhand how to factor $u^4+1$ into quadratics (probably from using the fourth roots of $-1$), but you should be able to use partial fractions from there.