Question

\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]

Answer 1

\[\text{We have}, \]

\[I = \int \sin^{- 1} \left( 3x - 4 x^3 \right)dx\]

\[\text{ Putting x }= \sin \theta \Rightarrow \theta = \sin^{- 1} x\]

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

\[ \therefore I = \int \sin^{- 1} \left( 3 \sin \theta - 4 \sin^3 \theta \right) \cos \text{  θ  dθ}\]

\[ = \int \sin^{- 1} \left( \sin 3\theta \right) \cos \text{  θ  dθ}\]

\[ = 3\int \theta_I \text{ cos}_{II} \text{  θ  dθ}\]

\[ = 3 \left[ \theta \left( \sin \theta \right) - \int1 \sin \text{  θ  dθ} \right]\]

\[ = 3\left[ \theta \sin \theta + \cos \theta \right] + C\]

\[ = 3\left[ \theta \sin \theta + \sqrt{1 - \sin^2 \theta} \right] + C\]

\[ = 3 \left[ \sin^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]

\[ = 3 \left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right] + C\]