The value of integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(sinx-xcosx)}{x(x+sinx)}dx$ is
The value of integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(sinx-xcosx)}{x(x+sinx)}dx$ is
- $\log _{e}\left\{\frac{2(\pi+3)}{(2 \pi+3 \sqrt{3})}\right\}$
- $\log _{e}\left\{\frac{\pi+3}{2(2 \pi+3 \sqrt{3})}\right\}$
- $\log _{e}\left\{\frac{2 \pi+3 \sqrt{3}}{2(\pi+3)}\right\}$
- $\log _{e}\left\{\frac{2(2 \pi+3 \sqrt{3})}{\pi+3}\right\}$
Solution
Let $l=\int_{\pi / 6}^{\pi / 3} \frac{(\sin x-x \cos x)}{x(x+\sin x)} d x$
$\Rightarrow \quad I=\int_{\pi / 6}^{\pi / 3} \frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)} d x$
$\Rightarrow \quad I=\int_{\pi / 6}^{\pi / 3}\left(\frac{1}{x}-\frac{1+\cos x}{x+\sin x}\right) d x$
$\Rightarrow \quad 1=[\log \times]_{\pi / 6}^{\pi / 3}-\int_{\pi / 6}^{\pi / 3} \frac{1+\cos x}{x+\sin x} \cdot d x$
$\left\{\begin{array}{l}\text { put } t=x+\sin x \\ dt=(1+\cos x) d x\end{array}\right.$
$\Rightarrow \quad I=\left(\log \frac{\pi}{3}-\log \frac{\pi}{6}\right)-\int\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right) \frac{d t}{t}$
$\Rightarrow \quad I=\log 2-\left[\log t \mid\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right.$
$\Rightarrow \quad 1=\log 2-\left[\log \left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)-\log \left(\frac{\pi}{6}+\frac{1}{2}\right)\right]$
$\Rightarrow \quad I=\log 2-\log \left(\frac{2 \pi+3 \sqrt{3}}{\pi+3}\right)$
$\left(\because \log m-\log n=\log \frac{m}{n}\right)$
$I=\log \left(\frac{2(\pi+3)}{2 \pi+3 \sqrt{3}}\right)$
$=\log \left(\frac{2 \pi+6}{2 \pi+3 \sqrt{3}}\right)$
Asked in: TEST SERIES MHT-CET Full Test 6
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