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प्रश्न
\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]
योग
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उत्तर
\[\text{We have}, \]
\[I = \int \sin^{- 1} \left( 3x - 4 x^3 \right)dx\]
\[\text{ Putting x }= \sin \theta \Rightarrow \theta = \sin^{- 1} x\]
\[ \Rightarrow dx = \cos \text{ θ dθ}\]
\[ \therefore I = \int \sin^{- 1} \left( 3 \sin \theta - 4 \sin^3 \theta \right) \cos \text{ θ dθ}\]
\[ = \int \sin^{- 1} \left( \sin 3\theta \right) \cos \text{ θ dθ}\]
\[ = 3\int \theta_I \text{ cos}_{II} \text{ θ dθ}\]
\[ = 3 \left[ \theta \left( \sin \theta \right) - \int1 \sin \text{ θ dθ} \right]\]
\[ = 3\left[ \theta \sin \theta + \cos \theta \right] + C\]
\[ = 3\left[ \theta \sin \theta + \sqrt{1 - \sin^2 \theta} \right] + C\]
\[ = 3 \left[ \sin^{- 1} x \times x + \sqrt{1 - x^2} \right] + C\]
\[ = 3 \left[ x \sin^{- 1} x + \sqrt{1 - x^2} \right] + C\]
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