Question

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

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

Answer 1

Answer: (A)

${\log }_e\left\{\frac{2(\pi +3)}{(2\pi +3\sqrt{3})}\right\}$

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x-x\cos x}{x(x+\sin x)}dx$

$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)}dx$

$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}(\frac{1}{x}-\frac{1+\cos x}{x+\sin x})dx$

$\Rightarrow I={\left[\log x\right]}_{\pi /6}^{\pi /3}-{\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1+\cos x}{x+\sin x}dx$

Put $t=x+\sin x$; $dt=(1+\cos x)dx$ in $2^{nd}$ term

$\Rightarrow I=(\log \frac{\pi }{3}-\log \frac{\pi }{6})-{\int }_{(\frac{\pi }{6}+\frac{1}{2})}^{(\frac{\pi }{3}+\frac{\sqrt{3}}{2})}\frac{dt}{t}$

$\Rightarrow I=\log 2-{\left[\log t\right]}_{(\frac{\pi }{6}+\frac{1}{2})}^{(\frac{\pi }{3}+\frac{\sqrt{3}}{2})}$

$\Rightarrow I=\log 2-[\log (\frac{\pi }{3}+\frac{\sqrt{3}}{2})-\log (\frac{\pi }{6}+\frac{1}{2})]$

$\Rightarrow I=\log 2-\log \left(\frac{2\pi +3\sqrt{3}}{\pi +3}\right)(\because \log m-\log n=\log \frac{m}{n})$

$I=\log \left(\frac{2(\pi +3)}{2\pi +3\sqrt{3}}\right)=\log \left(\frac{2\pi +6}{2\pi +3\sqrt{3}}\right)$