The value of int limits_{frac{pi}{3}}^{frac{p | vedclass

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Question

(C) Let $I = \int \limits_{\pi / 3}^{\pi / 2} \frac{2+3 \sin x}{\sin x(1+\cos x)} d x = 2 \int \limits_{\pi / 3}^{\pi / 2} \frac{d x}{\sin x(1+\cos x)

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$\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3}$

Vedclass · Answer

(C) Let $I = \int \limits_{\pi / 3}^{\pi / 2} \frac{2+3 \sin x}{\sin x(1+\cos x)} d x = 2 \int \limits_{\pi / 3}^{\pi / 2} \frac{d x}{\sin x(1+\cos x)} + 3 \int \limits_{\pi / 3}^{\pi / 2} \frac{d x}{1+\cos x}$.First,evaluate $I_1 = \int \limits_{\pi / 3}^{\pi / 2} \frac{d x}{1+\cos x} = \int \limits_{\pi / 3}^{\pi / 2} \frac{1-\cos x}{\sin^2 x} d x = \int \limits_{\pi / 3}^{\pi / 2} (\operatorname{cosec}^2 x - \cot x \operatorname{cosec} x) d x$.$I_1 = [-\cot x + \operatorname{cosec} x]_{\pi / 3}^{\pi / 2} = (0 + 1) - (-\frac{1}{\sqrt{3}} + \frac{2}{\sqrt{3}}) = 1 - \frac{1}{\sqrt{3}}$.Next,evaluate $I_2 = \int \limits_{\pi / 3}^{\pi / 2} \frac{d x}{\sin x(1+\cos x)}$. Using $t = 	an(x/2)$,$dx = \frac{2 dt}{1+t^2}$,$\sin x = \frac{2t}{1+t^2}$,$\cos x = \frac{1-t^2}{1+t^2}$.$I_2 = \int \limits_{1/\sqrt{3}}^{1} \frac{1}{\frac{2t}{1+t^2} (1 + \frac{1-t^2}{1+t^2})} \cdot \frac{2 dt}{1+t^2} = \int \limits_{1/\sqrt{3}}^{1} \frac{1+t^2}{2t} dt = \frac{1}{2} [\ln|t| + \frac{t^2}{2}]_{1/\sqrt{3}}^{1}$.$I_2 = \frac{1}{2} [(\ln 1 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})] = \frac{1}{2} [\frac{1}{3} + \ln \sqrt{3}] = \frac{1}{6} + \frac{1}{2} \ln \sqrt{3}$.Thus,$I = 2 I_2 + 3 I_1 = 2(\frac{1}{6} + \frac{1}{2} \ln \sqrt{3}) + 3(1 - \frac{1}{\sqrt{3}}) = \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3} = \frac{10}{3} - \sqrt{3} + \ln \sqrt{3}$.

The value of $\int \limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$ is equal to

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