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

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Question

(A) ધારો કે $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$ ગુણધર્મ $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ નો

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) ધારો કે $I = \int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x} \quad \dots (i)$ ગુણધર્મ $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ નો ઉપયોગ કરતા,આપણને મળે છે $I = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin(\pi-x)} = \int_{0}^{\pi} \frac{(\pi-x) \, dx}{1+\cos \alpha \sin x} \quad \dots (ii)$ $(i)$ અને $(ii)$ નો સરવાળો કરતા: $2I = \pi \int_{0}^{\pi} \frac{dx}{1+\cos \alpha \sin x}$ $\sin x = \frac{2 	an(x/2)}{1+	an^2(x/2)}$ અને $dx = \frac{2 \sec^2(x/2) \, d(x/2)}{1}$ નો ઉપયોગ કરતા,ધારો કે $t = 	an(x/2)$,તો $dt = \frac{1}{2} \sec^2(x/2) \, dx$: $2I = \pi \int_{0}^{\infty} \frac{2 \, dt}{1+t^2 + 2t \cos \alpha} = 2\pi \int_{0}^{\infty} \frac{dt}{(t+\cos \alpha)^2 + \sin^2 \alpha}$ $I = \pi \left[ \frac{1}{\sin \alpha} 	an^{-1} \left( \frac{t+\cos \alpha}{\sin \alpha} ight) ight]_{0}^{\infty}$ $I = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - 	an^{-1} \left( \cot \alpha ight) ight) = \frac{\pi}{\sin \alpha} \left( \frac{\pi}{2} - (\frac{\pi}{2} - \alpha) ight) = \frac{\pi \alpha}{\sin \alpha}$

$\int_{0}^{\pi} \frac{x \, dx}{1+\cos \alpha \sin x}, (0 < \alpha < \pi)$ ની કિંમત શોધો.

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