Advertisements
Advertisements
प्रश्न
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Advertisements
उत्तर
\[\int \cos^{-1} \left(\sin x \right)\text{ dx }\]
\[= \int \cos^{-1} \left( \cos\left( \frac{\pi}{2} - x \right) \right) \text{ dx }\]
\[ = \int \left(\frac{\pi}{2} - x \right) dx\]
\[ = \frac{\pi}{2}x - \frac{1}{2} x^2 + c\]
\[\text{Hence,}\int \cos^{-1} \left(\sin x \right)\text{ dx } = \frac{\pi}{2}x - \frac{1}{2} x^2 + c\]
संबंधित प्रश्न
` ∫ cot^3 x "cosec"^2 x dx `
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integrals:
Write a value of
Evaluate:
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{1}{x^2 + 16}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
