Advertisements
Advertisements
प्रश्न
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
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
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integrals:
Evaluate the following integral :-
Evaluate the following integral :-
Evaluate the following integral:
Evaluate the following integral:
Evaluate the following integral:
Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int\frac{\left( 1 + \log x \right)^2}{x} \text{ dx }\]
Evaluate:\[\int\frac{\log x}{x} \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
Evaluate the following:
`int ("d"x)/(xsqrt(x^4 - 1))` (Hint: Put x2 = sec θ)
Evaluate the following:
`int_1^2 ("d"x)/sqrt((x - 1)(2 - x))`
