A) \[\frac{x}{\sqrt{1-{{x}^{2}}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-{{x}^{2}})+c\]
B) \[\frac{x}{\sqrt{1-{{x}^{2}}}}{{\sin }^{-1}}x-\frac{1}{2}\log (1-{{x}^{2}})+c\]
C) \[\frac{1}{\sqrt{1-{{x}^{2}}}}{{\sin }^{-1}}x-\frac{1}{2}\log (1-{{x}^{2}})+c\]
D) \[\frac{1}{\sqrt{1-{{x}^{2}}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-{{x}^{2}})+c\]
Correct Answer: A
Solution :
Put \[t={{\sin }^{-1}}x\Rightarrow \sin t=x\Rightarrow \cos t\,dt=dx,\] then \[\int_{{}}^{{}}{\frac{{{\sin }^{-1}}x}{{{(1-{{x}^{2}})}^{3/2}}}\,dx}=\int_{{}}^{{}}{t{{\sec }^{2}}t\,dt=t\tan t+\log \cos t+c}\] \[={{\sin }^{-1}}x\tan ({{\sin }^{-1}}x)+\log \cos ({{\sin }^{-1}}x)+c\] \[=\frac{x}{\sqrt{1-{{x}^{2}}}}{{\sin }^{-1}}x+\frac{1}{2}\log (1-{{x}^{2}})+c.\]
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