A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}+c\]
B) \[x{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}+c\]
C) \[{{\cos }^{-1}}x+c\]
D) \[x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+c\]
Correct Answer: D
Solution :
\[I=\int_{{}}^{{}}{{{\sin }^{-1}}x}.1\,dx\,dx\] \[I={{\sin }^{-1}}x.x-\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{x}^{2}}}}}\,.\,x\,dx\] Put \[1-{{x}^{2}}={{t}^{2}}\Rightarrow -2xdx=2tdt\] in the second integral and solve it, therefore \[I=x{{\sin }^{-1}}x.+\sqrt{1-{{x}^{2}}}+c\].
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