int frac{dx}{cos 2x - cos^2 x} = | vedclass

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

Question

(D) We are given the integral $I = \int \frac{dx}{\cos 2x - \cos^2 x}$.Using the trigonometric identity $\cos 2x = 2\cos^2 x - 1$,we substitute this i

Vedclass · Accepted Answer

$\cot x + c$

Vedclass · Answer

(D) We are given the integral $I = \int \frac{dx}{\cos 2x - \cos^2 x}$.Using the trigonometric identity $\cos 2x = 2\cos^2 x - 1$,we substitute this into the denominator:$I = \int \frac{dx}{(2\cos^2 x - 1) - \cos^2 x} = \int \frac{dx}{\cos^2 x - 1}$.We know that $\cos^2 x - 1 = - (1 - \cos^2 x) = -\sin^2 x$.Substituting this back into the integral:$I = \int \frac{dx}{-\sin^2 x} = -\int \csc^2 x \, dx$.Since the derivative of $\cot x$ is $-\csc^2 x$,the integral of $-\csc^2 x$ is $\cot x + c$.Therefore,$I = \cot x + c$.

$\int \frac{dx}{\cos 2x - \cos^2 x} = $

Similar Questions

$\int a^x \, dx = $

Find the following integral: $\int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$

If $\int \frac{dx}{x^2+2x+2}=f(x)+c$,then $f(x)$ is equal to :

$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$

$\int \frac{1 + \tan x \tan(x + a)}{\tan x \tan(x + a)} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module