$\int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x$

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Solution

$\int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x Let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} d x = d t Now, \int \frac{\sin (\tan^{- 1} x)}{1 + x^{2}} d x = \int \sin t d t = - \cos (t) + C = - \cos (\tan^{- 1} x) + C$

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