If int frac{d x}{cot ^2 x-1}=frac{1}{A} log | | vedclass

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

Question

(B) We are given the integral $I = \int \frac{d x}{\cot ^2 x-1}$.First,rewrite the integrand using $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{d

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) We are given the integral $I = \int \frac{d x}{\cot ^2 x-1}$.First,rewrite the integrand using $\cot x = \frac{\cos x}{\sin x}$:$I = \int \frac{d x}{\frac{\cos ^2 x}{\sin ^2 x}-1} = \int \frac{\sin ^2 x}{\cos ^2 x-\sin ^2 x} d x$.Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$,we get:$I = \int \frac{\sin ^2 x}{\cos 2 x} d x$.Using the identity $\sin ^2 x = \frac{1-\cos 2 x}{2}$,we have:$I = \int \frac{1-\cos 2 x}{2 \cos 2 x} d x = \frac{1}{2} \int (\sec 2 x - 1) d x$.Integrating term by term:$I = \frac{1}{2} \left( \frac{\log |\sec 2 x + 	an 2 x|}{2} - x ight) + c$.$I = \frac{1}{4} \log |\sec 2 x + 	an 2 x| - \frac{x}{2} + c$.Comparing this with the given form $\frac{1}{A} \log |\sec 2 x + 	an 2 x| - \frac{x}{B} + c$,we find $A = 4$ and $B = 2$.Therefore,$A + B = 4 + 2 = 6$.

If $\int \frac{d x}{\cot ^2 x-1}=\frac{1}{A} \log |\sec 2 x+\tan 2 x|-\frac{x}{B}+c$,(where $c$ is the constant of integration),then $A+B=$

Similar Questions

Integrate the function: $\frac{x+3}{x^{2}-2x-5}$

If $\int (e^{2x} + 2e^{x} - e^{-x} - 1) e^{(e^{x} + e^{-x})} dx = g(x) e^{(e^{x} + e^{-x})} + c$,where $c$ is a constant of integration,then $g(0)$ is equal to

$\int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx = $

$\int \frac{dx}{(2\sin x + \cos x)^2} = $

If $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$, then the value of $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module