# Evaluate $\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx$

Evaluate
$$\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} dx.$$

Hint: First show that for any function $f(x)$ the following formula holds
$\int_{a}^{b}\frac{f(x)}{f(a+b-x)+f(x)}dx=\frac{b-a}{2}.$

Let $u=a+b-x$. Then $du=-dx$ and
$I=\int_a^{b}\frac{f(x)}{f(a+b-x)+f(x)}dx$
$=\int_{b}^{a}\frac{f(a+b-u)}{f(u)+f(a+b-u)}(-du)$
$=\int_{a}^{b}\frac{f(a+b-u)}{f(u)+f(a+b-u)}du$
$=\int_{a}^{b}\frac{f(a+b-x)}{f(x)+f(a+b-x)}dx=I.$
Hence
$2I=\int_a^{b}\frac{f(x)}{f(a+b-x)+f(x)}dx+\int_{a}^{b}\frac{f(a+b-x)}{f(x)+f(a+b-x)}dx$
$=\int_a^{b}\frac{f(x)+f(a+b-x)}{f(a+b-x)+f(x)}dx=\int_a^{b}1dx=b-a.$
So
$I=\int_a^{b}\frac{f(x)}{f(a+b-x)+f(x)}dx=\frac{b-a}{2}.$
Now letting $f(x)=\sqrt{\sin x}$, $a=0, b=\frac{\pi}{2}$. We have
$f(\frac{\pi}{2}+0-x)=\sqrt{\sin (\frac{\pi}{2}-x)}=\sqrt{\cos x}.$
Hence
$\int_0^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=\int_0^{\frac{\pi}{2}}\frac{f(x)}{f(0+\frac{\pi}{2}-x)+f(x)}dx$
$\frac{\frac{\pi}{2}-0}{2}=\frac{\pi}{4}.$

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