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प्रश्न
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx }\]
योग
उत्तर
\[\int\] sin–1 (3x – 4x3)dx
Let x = sin θ
⇒ dx = cos θ.dθ
& θ = sin–1 x
\[\int\] sin–1 (3x – 4x3)dx =
\[\int\] sin–1 (3 sin θ – 4 sin3 θ) . cos θ dθ
= ∫ sin–1 (sin 3θ) . cos θ dθ
\[= 3\int \theta_I . \cos _{II} \theta d\theta\]
\[ = 3\left[ \theta\int\cos \theta d\theta - \int\left\{ \frac{d}{d\theta}\left( \theta \right) - \int\cos \theta d\theta \right\}d\theta \right]\]
\[ = 3\left[ \theta . \sin \theta - \int1 . \sin \theta d\theta \right]\]
\[ = 3\left[ \theta . \sin \theta + \cos \theta \right] + C\]
\[ = 3\left[ \theta . \sin \theta + \sqrt{1 - \sin^2 \theta} \right] + C\]
\[ = 3\left[ \left( \sin^{- 1} x \right) . x + \sqrt{1 - x^2} \right] + C \left( \because \theta = \sin^{- 1} x \right)\]
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