A) \[\frac{1}{2}+\frac{\sqrt{3}\pi }{12}\]
B) \[\frac{1}{2}-\frac{\sqrt{3}\pi }{12}\]
C) \[\frac{1}{2}-\frac{\sqrt{3\pi }}{12}\]
D) None of these
Correct Answer: B
Solution :
Put \[t={{\sin }^{-1}}x\Rightarrow dt=\frac{1}{\sqrt{1-{{x}^{2}}}}dx,\] 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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