Question

The value of the integral is equal to $\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\left(sin\,x - xcos\, x\right)}{x\left(x+sin\,x\right)}dx$ is

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

Answer 1

Answer: (A)

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

Answer 2

Let l =\int_\limits{\pi / 6}^{\pi / 3} \frac{(\sin x-x \cos x)}{x(x+\sin x)} d x
\Rightarrow l=\int_\limits{\pi / 6}^{\pi / 3} \frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)} d x
\Rightarrow I=\int_\limits{\pi / 6}^{\pi / 3}\left(\frac{1}{x}-\frac{1+\cos x}{x+\sin x}\right) d x
\Rightarrow I=[\log x]_{\pi / 6}^{\pi / 3}-\int_\limits\limits{\pi / 6}^{\pi / 3} \frac{1+\cos x}{x+\sin x} \cdot d x
$\begin{cases}\text { put } t=x+\sin x \\\ d t=(1+\cos x) d x\end{cases}$ in lind term
$\Rightarrow I=\left(\log \frac{\pi}{3}-\log \frac{\pi}{6}\right)-\int_ {\left(\frac{\pi}{6}+\frac{1}{2}\right)}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right) \frac{d t}{t}$
$\Rightarrow I=\log 2-[\log t]\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)$
$\Rightarrow J=\log 2-\left[\log \left(\frac{\pi}{3}+\frac{1}{2}\right)\right.$
$\Rightarrow I=\log 2-\log \left(\frac{2 \pi+3 \sqrt{3}}{\pi+3}\right)$
$\left(\because \log m-\log n=\log \frac{m}{n}\right)$
$\Rightarrow=\log \left(\frac{2(\pi+3)}{2 \pi+3 \sqrt{3}}\right)$
$\Rightarrow=\log \left(\frac{2 \pi+6}{2 \pi+3 \sqrt{3}}\right)$