Question

${\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ is equal to

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. None of these

Answer 1

Answer: (C)

$\frac{\pi }{12}$

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

We know,

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

Using this property,

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

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

Add (1) and (3) equation

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

$I={\int }_{\pi /6}^{\pi /3}\frac{1}{2}dx$

$I=\frac{1}{2}{\left[x\right]}_{\pi /6}^{\pi /3}$

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