$\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}\,+\sqrt{\cos x}} \, dx$ How could you evaluate this integral?
$$\int_0^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}\,+\sqrt{\cos x}} \, dx$$
I think the answer is $\pi/4$.
 A: $$
\int_0^{\pi/2}\frac{\sqrt{\sin x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}} + \int_0^{\pi/2}\frac{\sqrt{\cos x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}} = \int_0^{\pi/2} 1\,dx = \frac \pi 2.
$$
If you can show that the two integrals are equal to each other, then it follows that each must be $\pi/4$.
The substitution $u = \pi/2-x$ and $du=-dx$, transforming $\displaystyle\int_0^{\pi/2}\cdots\,dx$ into $\displaystyle\int_{\pi/2}^0 \cdots\,(-du)$ will do it provided you recall trigonometric identities involving $\sin(\pi/2 - x)$ and $\cos(\pi/2-x)$.
A: HINT:
Evaluate $$\int_{a}^{b} \frac {f(x)}{f(x)+f(a+b-x)}dx.$$
A: Note that 
$$\int_0^{\pi/2}\frac{\sqrt{\sin x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}} = \int_0^{\frac \pi 2} \frac{\sqrt{\sin(\pi/2-x)}\,\,dx}{\sqrt{\sin(\pi/2-x)} + \sqrt{\cos(\pi/2-x)}} = \int_0^{\pi/2}\frac{\sqrt{\cos x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}} $$
Let the integral be $I$, then 
$$2I = \int_0^{\pi/2}\frac{\sqrt{\sin x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}} + \int_0^{\pi/2}\frac{\sqrt{\cos x} \,\, dx}{\sqrt{\sin x}\,+\sqrt{\cos x}}$$
You can do the rest.
A: Substitute $t=\frac {\pi}{2}-x$ then $t=x$ as it is a dummy variable.
Add this equation this to original integrand. It should be $2$ times the integral required after adding and you see the that the integrand is $1$ and you can easily integrate.
