Q49 / 120 · VITEEE 2017 · Mathematics
Value of $\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ is
- A$\frac{\pi}{2}$
- B$\frac{-\pi}{2}$
- Correct answerC$\frac{\pi}{4}$
- DNone of these
Solution
Let $\mathrm{I}=\int_{0}^{\pi / 2} \frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}} \mathrm{dx}......(i)$
Then, I=
$$
\begin{array}{l}
\int_{0}^{\pi / 2} \frac{\sqrt{\sin (\pi / 2-x)}}{\sqrt{\sin (\pi / 2-x)}+\sqrt{\cos (\pi / 2-x)}} d x \\
\Rightarrow \quad I=\int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x......(ii)
\end{array}
$$
Adding (i) and (ii), we get
2I
$$
\begin{array}{l}
\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x+\int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\
\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x=\int_{0}^{\pi / 2} 1 . d x=[x]_{0}^{\pi / 2}=\frac{\pi}{2}-0
\end{array}
$$
$$
\Rightarrow \quad \mathrm{I}=\frac{\pi}{4} \Rightarrow \int_{0}^{\pi / 2} \frac{\sqrt{\sin \mathrm{x}}}{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}} \mathrm{dx}=\frac{\pi}{4}
$$
Hence, the correct answer is option: 'C'.