Question

\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]

Answer 1

\[\int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \text{ dx }\]

\[\text{ Let x }= \tan \theta\]

\[dx = \text{ sec}^2  \text{  θ }\text{ dθ   }\]

\[ \therefore \int \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)dx = \int \sin^{- 1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right) . \sec^2  \text{ θ    dθ }\]

\[ = \int \sin^{- 1} \left( \sin 2\theta \right) . \sec^2 \text{ θ  dθ }\]

\[ = \int \left( 2\theta \right) \sec^2  \text{ θ  dθ }\]

\[ = 2\int \theta_I \sec^2_{II} \text{ θ  dθ }\]

\[ = 2\left[ \theta\int \sec^2 \text{ θ  dθ }- \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int s {ec}^2 \text{ θ  dθ } \right\}d\theta \right]\]

\[ = 2\left[ \theta . \tan \theta - \int1 . \tan \text{ θ  dθ }\right]\]

\[ = 2\left[ \theta \tan \theta - \text{ log }\left| \sec \theta \right| \right] + C\]

\[ = 2\left[ \theta \tan \theta - \text{ log }\left| 1 + \tan^2 \theta \right|^\frac{1}{2} \right] + C\]

\[ = 2\left[ \left( \tan^{- 1} x \right) \times x - \text{ log }\left( 1 + x^2 \right)^\frac{1}{2} \right] + C\]

\[ = 2 x \tan^{- 1} x - 2 \times \frac{1}{2}\text{ log}\left| 1 + x^2 \right| + C\]

\[ = 2 x \tan^{- 1} x - \text{ log }\left| 1 + x^2 \right| + C\]