Question

\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]

Answer 1

\[\text{ Let I } = \int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]

\[\text{ Putting  x } = \cos \theta\]

\[ \Rightarrow dx = - \text{ sin   θ   dθ}  \]

\[and\ \theta = \cos^{- 1} x\]

\[ \therefore I = \int \tan^{- 1} \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} \left( - \sin \theta \right) d\theta\]

\[ = \int \tan^{- 1} \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} \left( - \sin \theta \right) d\theta\]

\[ = \int \tan^{- 1} \left( \tan \frac{\theta}{2} \right) \left( - \sin \theta \right) d\theta\]

\[ = - \frac{1}{2}\int \theta_I \sin_{II} \theta   d\theta\]

\[ = - \frac{1}{2}\left[ \theta\int \sin\theta d\theta - \int\left\{ \left( \frac{d}{d\theta}\theta \right)\int\sin \theta d\theta \right\}d\theta \right]\]

\[ = - \frac{1}{2} \left[ \theta\left( - \cos \theta \right) - \int 1 . \left( - \cos \theta \right) d\theta \right]\]

\[ = - \frac{1}{2} \left[ - \theta \cos \theta + \sin \theta \right] + C\]

\[ = - \frac{1}{2} \left[ - \theta . \cos \theta + \sqrt{1 - \cos^2 \theta} \right] + C\]

\[ = - \frac{1}{2}\left[ - \cos^{- 1} x . x + \sqrt{1 - x^2} \right] + C \left[ \because \theta = \cos^{- 1} x \right]\]

\[ = \frac{x \cos^{- 1} x}{2} - \frac{\sqrt{1 - x^2}}{2} + C\]