Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{Let I} = \int\frac{{cosec}^2 x}{1 + \cot x}dx\]
\[\text{Putting}\ \text{cot x} = t\]
\[ \Rightarrow - {cosec}^2 x = \frac{dt}{dx}\]
\[ \Rightarrow {cosec}^\text{2} \text{ x dx }= - dt\]
\[ \therefore I = \int\frac{- dt}{1 + t}\]
\[ = - \text{ln} \left| 1 + t \right| + C\]
\[ = - \text{ln }\left| 1 + \cot x \right| + C \left[ \because t = \cot x \right]\]
APPEARS IN
संबंधित प्रश्न
Integrate the following w.r.t. x `(x^3-3x+1)/sqrt(1-x^2)`
Evaluate the following integrals:
` ∫ cot^3 x "cosec"^2 x dx `
\[\int\frac{\cot x}{\sqrt{\sin x}} dx\]
Evaluate the following integrals:
Evaluate the following integrals:
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:\[\int\frac{\sin \sqrt{x}}{\sqrt{x}} \text{ dx }\]
Evaluate:\[\int \sec^2 \left( 7 - 4x \right) \text{ dx }\]
Evaluate: \[\int 2^x \text{ dx }\]
Evaluate: \[\int\frac{x^3 - x^2 + x - 1}{x - 1} \text{ dx }\]
Evaluate:\[\int\frac{e\tan^{- 1} x}{1 + x^2} \text{ dx }\]
Evaluate: \[\int\frac{2}{1 - \cos2x}\text{ dx }\]
Evaluate:
Evaluate: `int_ (x + sin x)/(1 + cos x ) dx`
Evaluate the following:
`int ("d"x)/sqrt(16 - 9x^2)`
Evaluate the following:
`int sqrt(5 - 2x + x^2) "d"x`
Evaluate the following:
`int x/(x^4 - 1) "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))`
