MCQ
$\int_0^{1/2} {\frac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\,dx = } $
- A$\frac{1}{2} + \frac{{\sqrt 3 \pi }}{{12}}$
- ✓$\frac{1}{2} - \frac{{\sqrt 3 \pi }}{{12}}$
- C$\frac{1}{2} \pm \frac{{\sqrt {3\pi } }}{{12}}$
- DNone of these
$\Rightarrow dt = \frac{1}{{\sqrt {1 - {x^2}} }}dx,$ then
$\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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