Question

Evaluate ${\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\frac{(\surd \sin x)}{(\surd \sin x+\surd \cos x)}\right]dx$

• A. $\frac{\mathrm{\pi }}{4}$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Let $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ……(1)

$={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin \left(\frac{\mathrm{\pi }}{2}-x\right)}}{\sqrt{\sin \left(\frac{\mathrm{\pi }}{2}-x\right)}+\sqrt{\cos \left(\frac{\mathrm{\pi }}{2}-x\right)}}dx$ $\left[\mathbf{\because }{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{d}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathit{f}\mathbf{(}\mathit{a}\mathbf{+}\mathit{b}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{d}\mathit{x}\right]$

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ …….(2) $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\left(\frac{\pi }{2}-\theta \right)\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\frac{\pi }{2}-\theta \right)\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{\theta }\right]$

By adding equation (1) and (ii), we get

$\Rightarrow$$2I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$

$\Rightarrow$$2I={\int }_0^{\frac{\mathrm{\pi }}{2}}1dx$

$\Rightarrow$$2I={\left[x\right]}_0^{\frac{\mathrm{\pi }}{2}}$

$\Rightarrow$$2I=\frac{\mathrm{\pi }}{2}$

$\Rightarrow$ $I=\frac{\mathbf{\pi }}{\mathbf{4}}$

Hence, Option ‘A’ is Correct.