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