Question

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

Answer 1

\[\text{We have}, \]

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

\[\text{ Putting  x }= \cos \theta\]

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

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

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

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

\[ = - \frac{1}{2}\int\theta \sin \theta d\theta\]

\[\text{Considering} \text{  θ  as first function and} \sin \text{  θ   as second function}\]

\[I = - \frac{1}{2}\left[ \theta\left( - \cos \theta \right) - \int1\left( - \cos \theta \right)d\theta \right]\]

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

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

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

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

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