Question
$\int_0^{1/2} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\,dx = } $
$dt = \frac{1}{{\sqrt {1 - {x^2}} }}dx$
अब $\int_0^{1/2} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx = \int_0^{\pi /6} {\,\,t\sin t\,dt} } $
$ = [ - t\cos t + \sin t]_0^{\pi /6}$
$ = \left[ { - \frac{\Pi }{6}.\frac{{\sqrt 3 }}{2} + \frac{1}{2}} \right] $
$= \left[ {\frac{1}{2} - \frac{{\sqrt 3 \Pi }}{{12}}} \right]$
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