The value of the integral int_{frac{pi}{6}}^{ | vedclass

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(A) Let $I = \int_{\pi/6}^{\pi/3} \frac{\sin x - x \cos x}{x(x + \sin x)} dx$We can rewrite the numerator as $(x + \sin x) - x(1 + \cos x)$.So,$I = \i

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$\log_{e} \left\{ \frac{2(\pi + 3)}{(2\pi + 3\sqrt{3})} \right\}$

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(A) Let $I = \int_{\pi/6}^{\pi/3} \frac{\sin x - x \cos x}{x(x + \sin x)} dx$We can rewrite the numerator as $(x + \sin x) - x(1 + \cos x)$.So,$I = \int_{\pi/6}^{\pi/3} \left( \frac{x + \sin x}{x(x + \sin x)} - \frac{x(1 + \cos x)}{x(x + \sin x)} \right) dx$$I = \int_{\pi/6}^{\pi/3} \left( \frac{1}{x} - \frac{1 + \cos x}{x + \sin x} \right) dx$Integrating term by term: $I = [\log |x|]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} \frac{1 + \cos x}{x + \sin x} dx$For the second integral,let $t = x + \sin x$,then $dt = (1 + \cos x) dx$.When $x = \pi/6$,$t = \pi/6 + 1/2 = (\pi + 3)/6$. When $x = \pi/3$,$t = \pi/3 + \sqrt{3}/2 = (2\pi + 3\sqrt{3})/6$.$I = \log(\pi/3) - \log(\pi/6) - [\log |t|]_{(\pi+3)/6}^{(2\pi+3\sqrt{3})/6}$$I = \log(2) - \left( \log \left( \frac{2\pi + 3\sqrt{3}}{6} \right) - \log \left( \frac{\pi + 3}{6} \right) \right)$$I = \log(2) - \log \left( \frac{2\pi + 3\sqrt{3}}{\pi + 3} \right) = \log \left( \frac{2(\pi + 3)}{2\pi + 3\sqrt{3}} \right)$

The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x - x \cos x)}{x(x + \sin x)} dx$ is

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