Q67 / 120 · VITEEE 2019 · Mathematics
Value of $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$ is
- A$\frac{\pi}{6}$
- Correct answerB$\frac{\pi}{12}$
- C$\frac{12}{\pi}$
- DNone of these
Solution
$\begin{aligned} \mathrm{I} &=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} \mathrm{dx} \\ &=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} \mathrm{dx} \end{aligned} ......(i)$
Then, $\mathrm{I}=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin \left(\frac{\pi}{2}-\mathrm{x}\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-\mathrm{x}\right)}+\sqrt{\cos \left(\frac{\pi}{2}-\mathrm{x}\right)}} \mathrm{dx}$
$\Rightarrow \mathrm{I}=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}} \mathrm{dx} ......(ii)$
Adding (i) and (ii), we get
$$
\begin{array}{l}
2 \mathrm{I}=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin \mathrm{x}}+\sqrt{\cos \mathrm{x}}}{\sqrt{\cos \mathrm{x}}+\sqrt{\sin \mathrm{x}}} \mathrm{dx} \\
\Rightarrow 2 \mathrm{I}=\int_{\pi / 6}^{\pi / 3} 1 \cdot \mathrm{dx}=[\mathrm{x}]_{\pi / 6}^{\pi / 3}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \Rightarrow \mathrm{I}=\pi / 12
\end{array}
$$
Hence, the correct answer is option: 'B'.