int_{pi / 6}^{pi / 3} frac{1}{1+sqrt{cot x}} | vedclass

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Question

(A) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$. We can rewrite the integrand as $I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x

Vedclass · Accepted Answer

$\frac{\pi}{12}$

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(A) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$. We can rewrite the integrand as $I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ $(i)$. Using the property $\int_a^b f(x) d x = \int_a^b f(a+b-x) d x$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we get: $I = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x = \int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$ (ii). Adding $(i)$ and (ii): $2I = \int_{\pi / 6}^{\pi / 3} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) d x = \int_{\pi / 6}^{\pi / 3} 1 d x$. $2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Therefore,$I = \frac{\pi}{12}$.

$\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=$

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