int_{0}^{pi} frac{x , dx}{1+cos alpha sin x}, | vedclass

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(A) Let $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$ Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \,

Vedclass · Accepted Answer

$\frac{\pi \alpha}{\sin \alpha}$

Vedclass · Answer

(A) Let $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$ Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$,we get $I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin(\pi-x)} = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin x} \quad \dots (ii)$ Adding $(i)$ and $(ii)$: $2I = \pi \int_{0}^{\pi} \frac{dx}{1+\cos \alpha \sin x}$ Using $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$ and $dx = \frac{2 \sec^2(x/2) \, d(x/2)}{1}$,let $t = 	an(x/2)$,then $dt = \frac{1}{2} \sec^2(x/2) \, dx$: $2I = \pi \int_{0}^{\infty} \frac{2 \, dt}{1+t^2 + 2t \cos \alpha} = 2\pi \int_{0}^{\infty} \frac{dt}{(t+\cos \alpha)^2 + \sin^2 \alpha}$ $I = \pi \left[ \frac{1}{\sin \alpha} 	an^{-1} \left( \frac{t+\cos \alpha}{\sin \alpha} ight) ight]_{0}^{\infty}$ $I = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - 	an^{-1} \left( \cot \alpha ight) ight) = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - (\frac{\pi}{2} - \alpha) ight) = \frac{\pi \alpha}{\sin \alpha}$

$\int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x}, (0 < \alpha < \pi)$ is equal to

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