Advertisements
Advertisements
प्रश्न
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Advertisements
उत्तर
\[\int\frac{2}{1 - \cos2x}dx = \int\frac{2}{2 \sin^2 x}dx\]
\[ = \int {cosec}^2 \text{ x dx }\]
\[ = - \cot x + c\]
\[\text{ Hence, } \int\frac{2}{1 - \cos2x}dx = - \cot x + c .\]
APPEARS IN
संबंधित प्रश्न
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
` ∫ cot^3 x "cosec"^2 x dx `
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:
Write a value of
Write a value of
Evaluate:\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{1}{\sqrt{1 - x^2}} \text{ dx }\]
Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{ dx }\]
Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]
Evaluate:
\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]
Evaluate:
`∫ (1)/(sin^2 x cos^2 x) dx`
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Evaluate the following:
`int (3x - 1)/sqrt(x^2 + 9) "d"x`
Evaluate the following:
`int sqrt(2"a"x - x^2) "d"x`
Evaluate the following:
`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`
