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