$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=$
$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=$
- $\frac{\pi}{12}$
- $\frac{\pi}{6}$
- $\frac{\pi}{4}$
- $\frac{\pi}{13}$
Solution
$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=I$
$
I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x
$
We have, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
$
\begin{gathered}
a+b-x=\frac{\pi}{2}-x \\
I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\
I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x
\end{gathered}
$
Eqs. (i) + (ii) $\Rightarrow 2 I$
$
\begin{aligned}
& =\int_{\pi / 6}^{\pi / 3}\left(\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right) d x \\
& =\int_{\pi / 6}^{\pi / 3}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right) d x \\
& =\int_{\pi / 6}^{\pi / 3} 1 \cdot d x=(x)_{\pi / 6}^{\pi / 3}=\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \\
2 I & =\frac{\pi}{6} \Rightarrow I=\frac{\pi}{12}
\end{aligned}
$
Hence, option (1) is correct
Asked in: AP EAMCET 2020 (22 Sep Shift 2)
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