A) \[{{\cos }^{-1}}(x-1)+c\]
B) \[{{\sin }^{-1}}(x-1)+c\]
C) \[{{\cos }^{-1}}(1+x)+c\]
D) \[{{\sin }^{-1}}(1-x)+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{\frac{dx}{\sqrt{2x-{{x}^{2}}}}}=\int_{{}}^{{}}{\frac{dx}{\sqrt{1-{{(x-1)}^{2}}}}}={{\sin }^{-1}}(x-1)+c.\]
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