int_{frac{pi}{6}}^{frac{pi}{2}} frac{operator | vedclass

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(D) ધારો કે $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x$. $t = \operatorname{

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$	an ^{-1}\left(\frac{1}{3}ight)$

Vedclass · Answer

(D) ધારો કે $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x$. $t = \operatorname{cosec} x$ આદેશ લેતા,$dt = -\operatorname{cosec} x \cot x \, dx$ મળે,તેથી $\operatorname{cosec} x \cot x \, dx = -dt$. જ્યારે $x = \frac{\pi}{6}$,ત્યારે $t = \operatorname{cosec}(\frac{\pi}{6}) = 2$. જ્યારે $x = \frac{\pi}{2}$,ત્યારે $t = \operatorname{cosec}(\frac{\pi}{2}) = 1$. તેથી,$I = \int_{2}^{1} \frac{-dt}{1+t^2} = \int_{1}^{2} \frac{dt}{1+t^2}$. સંકલન કરતા,$I = [	an^{-1} t]_{1}^{2} = 	an^{-1}(2) - 	an^{-1}(1)$. સૂત્ર $	an^{-1} x - 	an^{-1} y = 	an^{-1}(\frac{x-y}{1+xy})$ નો ઉપયોગ કરતા,$I = 	an^{-1}(\frac{2-1}{1+2 \cdot 1}) = 	an^{-1}(\frac{1}{3})$.

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\operatorname{cosec} x \cdot \cot x}{1+\operatorname{cosec}^2 x} d x=$

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