int_{frac{pi}{6}}^{frac{pi}{3}} frac{sqrt{cos | vedclass

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(D) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$. गुणधर्म $\int_{a}^{b} f(x) d x = \int_{a}^{

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$\frac{\pi}{12}$

Vedclass · Answer

(D) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$. गुणधर्म $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$ का उपयोग करते हुए,जहाँ $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. अतः,$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}} d x$. $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$. $I$ के दोनों समीकरणों को जोड़ने पर: $2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 d x$. $2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. अतः,$I = \frac{\pi}{12}$.

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$ का मान . . . . . . है।

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