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Evaluate: dx∫cos−1⁡(sin⁡x)dx - Mathematics

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प्रश्न

Evaluate:

\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]

योग
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उत्तर

\[\int \cos^{-1} \left(\sin x \right)\text{ dx }\]

\[= \int \cos^{-1} \left( \cos\left( \frac{\pi}{2} - x \right) \right) \text{ dx }\]

\[ = \int \left(\frac{\pi}{2} - x \right) dx\]

\[ = \frac{\pi}{2}x - \frac{1}{2} x^2 + c\]

\[\text{Hence,}\int \cos^{-1} \left(\sin x \right)\text{ dx } = \frac{\pi}{2}x - \frac{1}{2} x^2 + c\]

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अध्याय 19: Indefinite Integrals - Very Short Answers [पृष्ठ १९८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Very Short Answers | Q 60 | पृष्ठ १९८

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