A) \[x-{{\sin }^{-1}}x+C\]
B) \[x-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+C\]
C) \[x+{{\sin }^{-1}}x+C\]
D) \[x+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+C\]
E) \[x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C\]
Correct Answer: B
Solution :
\[\int{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx\] Put \[t={{\sin }^{-1}}x\] \[\Rightarrow \] \[x=\sin t\] \[dt=\frac{dx}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \] \[\int{x\,.\,tdt}\] \[\int{\underset{I}{\mathop{t}}\,.\underset{II}{\mathop{\sin }}\,t\,dt}\] \[=[t(-\cos t)-\int{1.(-\cos t)dt]}\] \[=-t\cos t+\sin t+C\] \[=-t\sqrt{1-{{\sin }^{2}}t}+\sin t+C\] \[=-\sqrt{1-{{x}^{2}}}.{{\sin }^{-1}}x+x+C\] \[=x-\sqrt{1-{{x}^{2}}}.{{\sin }^{-1}}x+C\]
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