int_{,pi /6}^{,pi /3} {,frac{{dx}}{{1 + sqrt | vedclass

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Question

(C) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\pi

Vedclass · Accepted Answer

$\pi /12$

Vedclass · Answer

(C) Let $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.We can write $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,so $I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ....$(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a = \pi/6$ and $b = \pi/3$,we have $a+b = \pi/2$.Thus,$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ....(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}} ight) dx = \int_{\pi /6}^{\pi /3} 1 dx$$2I = [x]_{\pi /6}^{\pi /3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$Therefore,$I = \frac{1}{2} 	imes \frac{\pi}{6} = \frac{\pi}{12}$.

$\int_{\,\pi /6}^{\,\pi /3} {\,\frac{{dx}}{{1 + \sqrt {\cot x} }}} $ is equal to:

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