Question

Evaluate :
${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

Answer 1

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$........(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$........(2)

Since $\sin (\frac{\pi }{2}-\theta )=\cos \theta$

Now add (1) and (2)

$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

$\therefore I=\frac{\pi }{12}$