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प्रश्न
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2} dx\]
बेरीज
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उत्तर
\[\int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2}dx\]
\[\text{Let} \tan^{- 1} x = t\]
\[ \Rightarrow \frac{1}{1 + x^2}dx = dt\]
\[Now, \int\frac{\sin \left( \tan^{- 1} x \right)}{1 + x^2}dx\]
\[ = \int\text{sin t dt} \]
\[ = - \cos \left( t \right) + C\]
\[ = - \cos \left( \tan^{- 1} x \right) + C\]
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