A) \[x-\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+c\]
B) \[x+\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x+c\]
C) \[\sqrt{1-{{x}^{2}}}{{\sin }^{-1}}x-x+c\]
D) None of these
Correct Answer: A
Solution :
Putting \[{{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}\,dx=dt,\] we get \[\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\,dx=}\int_{{}}^{{}}{t\sin t\,dt=-t\cos t+\sin t+c}\] \[=-{{\sin }^{-1}}x\cos ({{\sin }^{-1}}x)+\sin ({{\sin }^{-1}}x)+c\] \[=x-{{\sin }^{-1}}x\sqrt{1-{{x}^{2}}}+c.\]
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