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

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Question

(A) माना $I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\cot x}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.निश्चित

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(A) माना $I = \int_{\pi/6}^{\pi/3} \frac{1}{1+\sqrt{\cot x}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$.निश्चित समाकलन के गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करने पर,जहाँ $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$ है।अतः,$I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin(\pi/2-x)}}{\sqrt{\sin(\pi/2-x)} + \sqrt{\cos(\pi/2-x)}} dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$.$I$ के दोनों व्यंजकों को जोड़ने पर:$2I = \int_{\pi/6}^{\pi/3} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) dx = \int_{\pi/6}^{\pi/3} 1 dx$.$2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.इसलिए,$I = \frac{\pi}{12}$.

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

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