Integrate the function ^ - 1 ( 2x 1 + x^2 ) - Vidyadip

Question

Integrate the function ${\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$

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Answer

Putting $x = \tan \theta $
$ \Rightarrow dx = {\sec ^2}\theta d\theta $
$\therefore \int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} $
$= \int {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right).{{\sec }^2}\theta d\theta }$
$= \int {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right).{{\sec }^2}\theta d\theta } $
$= \int {2\theta {{\sec }^2}\theta d\theta }$
$ = 2\int {\theta {{\sec }^2}\theta d\theta } $ ...[Applying product rule]
$= 2\left[ {\theta .\tan \theta - \int {1.\tan \theta d\theta } } \right]$
$= 2\left[ {\theta .\tan \theta - \int {\tan \theta d\theta } } \right]$
$ = 2\left[ {\theta \tan \theta - \log \sec \theta } \right] + c$
$= 2\left[ {{{\tan }^{ - 1}}x.x - \log \sqrt {1 + {x^2}} } \right] + c$
$= 2\left[ {x{{\tan }^{ - 1}}x - \frac{1}{2}\log \left( {1 + {x^2}} \right)} \right] + c$
$= 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + c$

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