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