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प्रश्न
\[\int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
बेरीज
उत्तर
\[\text{ Let I} = \int \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) \text{ dx }\]
\[ = 2 \int 1_{II} . \tan^{- 1}_I \text{ x dx } \left[ \because \cos {}^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) = 2 \tan^{- 1} x \right] \]
\[ = 2\left[ \tan^{- 1} x\int1\text{ dx } - \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int1 \text{ dx }\right\}dx \right]\]
\[ = 2\left[ \tan^{- 1} x . x - \int\frac{1}{1 + x^2} \times\text{ x dx } \right]\]
\[ = 2 x \tan^{- 1} x - \int\frac{2x}{1 + x^2} \text{ dx }\]
\[\text{ Putting 1 }+ x^2 = t\]
\[ \Rightarrow \text{ 2x dx } = dt\]
\[ \therefore I = 2x \tan^{- 1} x - \int \frac{dt}{t}\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| t \right| + C\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| 1 + x^2 \right| + C \left[ \because t = 1 + x^2 \right]\]
\[ = 2 \int 1_{II} . \tan^{- 1}_I \text{ x dx } \left[ \because \cos {}^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) = 2 \tan^{- 1} x \right] \]
\[ = 2\left[ \tan^{- 1} x\int1\text{ dx } - \int\left\{ \frac{d}{dx}\left( \tan^{- 1} x \right)\int1 \text{ dx }\right\}dx \right]\]
\[ = 2\left[ \tan^{- 1} x . x - \int\frac{1}{1 + x^2} \times\text{ x dx } \right]\]
\[ = 2 x \tan^{- 1} x - \int\frac{2x}{1 + x^2} \text{ dx }\]
\[\text{ Putting 1 }+ x^2 = t\]
\[ \Rightarrow \text{ 2x dx } = dt\]
\[ \therefore I = 2x \tan^{- 1} x - \int \frac{dt}{t}\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| t \right| + C\]
\[ = 2x \tan^{- 1} x - \text{ ln }\left| 1 + x^2 \right| + C \left[ \because t = 1 + x^2 \right]\]
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