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प्रश्न
उत्तर
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right)\text{ dx }\]
\[\text{ Let x } = \cos \theta \]
\[ \Rightarrow \theta = \cos^{- 1} x\]
\[\text{and}\ dx = - \sin \text{ θ dθ }\]
\[ \therefore \int \cos^{- 1} \left( 4 x^3 - 3x \right)dx = \int \cos^{- 1} \left( 4 \cos^3 \theta - 3 \cos \theta \right) . \left( - \sin \theta \right)d\theta\]
\[ = \int \cos^{- 1} \left( \text{ cos 3 }\theta \right) . \left( - \sin \theta \right)d\theta \left( \because \text{ cos } \text{ 3 θ }= 4 \cos^3 \theta - 3 \cos \theta \right)\]
\[ = - 3 \int \theta_I \sin_{II} \text{ θ dθ }\]
\[ = \theta\int\sin \text{ θ dθ } - \int\left\{ \frac{d}{d\theta}\left( \theta \right)\int\sin \text{ θ dθ }\right\}d\theta\]
\[ = 3 \left[ \theta \left( - \cos \theta \right) - \int1 . \left( - \cos \theta \right)d\theta \right]\]
\[ = 3\theta \cos \theta - 3 \sin \theta + C \]
\[ = 3 \cos^{- 1} x . x - 3\sqrt{1 - x^2} + C \left( \because x = \cos \theta \right)\]
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