Question

$ \int{\tan ({{\sin }^{-1}}x)}dx $ is equal to

• 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 $

Answer 1

Answer: (E)

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 $