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

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(C) माना $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.हम जानते हैं कि $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,अतः $I = \int

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$\pi /12$

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(C) माना $I = \int_{\pi /6}^{\pi /3} \frac{dx}{1 + \sqrt{\cot x}}$.हम जानते हैं कि $\sqrt{\cot x} = \frac{\sqrt{\cos x}}{\sqrt{\sin x}}$,अतः $I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ ....$(i)$गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करने पर,जहाँ $a = \pi/6$ और $b = \pi/3$,इसलिए $a+b = \pi/2$.अतः,$I = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx = \int_{\pi /6}^{\pi /3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ ....(ii)$(i)$ और (ii) को जोड़ने पर:$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}$इसलिए,$I = \frac{1}{2} 	imes \frac{\pi}{6} = \frac{\pi}{12}$.

$\int_{\,\pi /6}^{\,\pi /3} {\,\frac{{dx}}{{1 + \sqrt {\cot x} }}} $ का मान ज्ञात कीजिए।

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