Question

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

Answer 1

\[\int\]  sin^–1 (3x – 4x³)dx
Let x = sin θ
⇒ dx = cos​ θ.d​θ
& θ = sin^–1 x

\[\int\]  sin^–1 (3x – 4x³)dx =

\[\int\]  sin^–1 (3 sin ​θ – 4 sin³ ​θ) . cos ​θ d​θ

                                = ∫ sin^–1 (sin 3​θ) . cos ​θ d​θ

\[= 3\int \theta_I . \cos _{II} \theta   d\theta\]

\[ = 3\left[ \theta\int\cos \theta d\theta - \int\left\{ \frac{d}{d\theta}\left( \theta \right) - \int\cos \theta d\theta \right\}d\theta \right]\]

\[ = 3\left[ \theta . \sin \theta - \int1 . \sin \theta d\theta \right]\]

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

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

\[ = 3\left[ \left( \sin^{- 1} x \right) . x + \sqrt{1 - x^2} \right] + C \left( \because \theta = \sin^{- 1} x \right)\]