Question

The value of the integral ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(sinx-xcosx)}{x(x+sinx)}dx$ is equal to

• A. $lo{g}_e\left(\frac{2(\pi +3)}{2\pi +3\sqrt{3}}\right)$
• B. $lo{g}_e\left(\frac{\pi +3}{2(2\pi +3\sqrt{3})}\right)$
• C. $lo{g}_e\left(\frac{2\pi +3\sqrt{3}}{2(\pi +3)}\right)$
• D. $lo{g}_e\left(\frac{2(2\pi +3\sqrt{3})}{\pi +3}\right)$

Answer 1

The correct option is A $lo{g}_e\left(\frac{2(\pi +3)}{2\pi +3\sqrt{3}}\right)$

Given : ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x-x\cos x)dx}{x(x+\sin x)}$

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x-x\cos x)dx}{x(x+\sin x)}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(\sin x-x\cos x)dx}{x^2(1+\frac{\sin x}{x})}Lett=1+\frac{\sin x}{x}x\to \frac{\pi }{6}t=1+\frac{3}{\pi }=\frac{\pi +3}{\pi }dt=\frac{x\cos x-\sin xdx}{x^2}x\to \frac{\pi }{3}t=1+\frac{3\sqrt{3}}{2\pi }=\frac{2\pi +3\sqrt{3}}{2\pi }I={\int }_{\frac{\pi +3}{\pi }}^{\frac{2\pi +3\sqrt{3}}{2\pi }}-\frac{dt}{t}=-\ln t=\ln {\left|\frac{1}{t}\right|}_{\frac{\pi +3}{\pi }}^{\frac{2\pi +3\sqrt{3}}{2\pi }}=\ln \left|\frac{2\pi +3\sqrt{3}}{2\pi }\right|-\ln \left|\frac{\pi +3}{\pi }\right|=\ln \left|\frac{2(\pi +3)}{2\pi +3\sqrt{3}}\right|$