int frac{csc^2 x}{1 + cot x} dx = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int \frac{\csc^2 x}{1 + \cot x} dx$. Substitute $1 + \cot x = t$. Differentiating both sides with respect to $x$,we get $-\csc^2 x dx =

Vedclass · Accepted Answer

$-\log(1 + \cot x) + c$

Vedclass · Answer

(B) Let $I = \int \frac{\csc^2 x}{1 + \cot x} dx$. Substitute $1 + \cot x = t$. Differentiating both sides with respect to $x$,we get $-\csc^2 x dx = dt$,which implies $\csc^2 x dx = -dt$. Substituting these into the integral,we get $I = \int \frac{-dt}{t}$. Integrating,we get $I = -\log|t| + c$. Replacing $t$ with $1 + \cot x$,we get $I = -\log|1 + \cot x| + c$.

$\int \frac{\csc^2 x}{1 + \cot x} dx = $

Similar Questions

$\int \frac{x}{1+x^4} \, dx =$

$\int \frac{e^{\tan^{-1} x}}{1 + x^2} dx = $

$\int \frac{1}{3-2 \cos 2 x} \,d x=$ (where $C$ is constant of integration.)

If $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 1 + 2x^7)^2} dx, x \geq 0$ and $f(0) = 0$,then the value of $f(1)$ is

$\int \frac{\sec^2 x}{(\sec x + \tan x)^2} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module