Question

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

Answer 1

\[\text{We have}, \]

\[I = \int \cos^{- 1} \left( 1 - 2 x^2 \right)dx\]

\[\text{ Putting x }= \sin \theta\]

\[ \Rightarrow dx = \cos \text{ θ   dθ}\]

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

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

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

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

\[ = 2\left[ \theta \sin \theta + \cos \theta \right] + C\]

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

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