=∫0π2 tsin⁡tcos⁡tdt=∫0π2 t⋅cot⁡tdt put sin−1⁡x=t x=sin⁡tdx=cos⁡tdt t=sin−1⁡0 , t=sin−1(1) t=0 , t=π2 =(tlog⁡|sin⁡t|]0π2−∫0π2 log⁡|sin⁡t|dt=(0−0)−−π2⋅log⁡2=π2⋅log⁡2

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$\int_{0}^{1} \frac{\sin^{- 1} x}{x} d x$

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$π \log 2$

$\frac{π}{2} \log 2$

$- \frac{π}{2} \log 2$

$- π \log 2$

answer is D.

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Detailed Solution

$= \int_{0}^{\frac{π}{2}} \frac{t}{\sin t} \cos t d t = \int_{0}^{\frac{π}{2}} t \cdot \cot t d t put \sin^{- 1} x = t x = \sin t d x = \cos t d t$

$t = \sin^{- 1} 0, t = \sin^{- 1} (1)$

$t = 0, t = \frac{π}{2}$

$\begin{aligned} = (t \log | \sin t |]_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} \log | \sin t | d t \\ = (0 - 0) - [- \frac{π}{2} \cdot \log 2] = \frac{π}{2} \cdot \log 2 \end{aligned}$