Question

Evaluate each of the following integrals:

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

Answer 1

Let I = ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ .....(1)
Then,
$$\begin{gathered} I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{3}}+{\displaystyle \frac{\mathrm{\pi }}{6}}-x\right)}}{\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{3}}+{\displaystyle \frac{\mathrm{\pi }}{6}}-x\right)}+\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}+{\displaystyle \frac{\mathrm{\pi }}{6}}-x\right)}}dx\left[{\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx\right] \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}{\sqrt{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}+\sqrt{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}}dx \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx.....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get

$$\begin{gathered} 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx \\ \Rightarrow 2I={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}}dx \\ \Rightarrow 2I={\overline{)x}}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{3}} \\ \Rightarrow 2I=\frac{\mathrm{\pi }}{3}-\frac{\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow I=\frac{\mathrm{\pi }}{12} \end{gathered}$$