A) \[\frac{1}{\sqrt{1-{{x}^{2}}}}+c\]
B) \[\sqrt{1-{{x}^{2}}}+c\]
C) \[\frac{-x}{\sqrt{1-{{x}^{2}}}}+c\]
D) \[\frac{x}{\sqrt{1-{{x}^{2}}}}+c\]
E) \[-\sqrt{1-{{x}^{2}}}+c\]
Correct Answer: E
Solution :
Let \[I=\int{\tan ({{\sin }^{-1}}x)}dx\] \[=\int{\tan \left( {{\tan }^{-1}}\frac{x}{\sqrt{1-{{x}^{2}}}} \right)}dx\] \[=\int{\frac{x}{\sqrt{1-{{x}^{2}}}}}dx\] Put \[1-{{x}^{2}}={{t}^{2}}\Rightarrow -2xdx=2t\,dt\] \[\therefore \]\[I=-\int{\frac{t\,dt}{t}}=-t+c\] \[\Rightarrow \] \[I=-\sqrt{1-{{x}^{2}}}+c\]
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