int {frac{{cos 2x - cos 2alpha }}{{cos x - co | vedclass

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

Question

(A) We are given the integral $I = \int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} \,dx$.Using the trigonometric identity $\cos 2	he

Vedclass · Accepted Answer

$2(\sin x + x\cos \alpha ) + c$

Vedclass · Answer

(A) We are given the integral $I = \int {\frac{{\cos 2x - \cos 2\alpha }}{{\cos x - \cos \alpha }}} \,dx$.Using the trigonometric identity $\cos 2	heta = 2\cos^2 	heta - 1$,we can rewrite the numerator:$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1) = 2\cos^2 x - 2\cos^2 \alpha = 2(\cos^2 x - \cos^2 \alpha)$.Now,substitute this into the integral:$I = \int {\frac{{2(\cos^2 x - \cos^2 \alpha )}}{{\cos x - \cos \alpha }}} \,dx$.Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$,we have:$I = \int {\frac{{2(\cos x - \cos \alpha )(\cos x + \cos \alpha )}}{{\cos x - \cos \alpha }}} \,dx$.Canceling the common term $(\cos x - \cos \alpha)$,we get:$I = 2\int {(\cos x + \cos \alpha )} \,dx$.Since $\cos \alpha$ is a constant with respect to $x$,we integrate term by term:$I = 2(\sin x + x\cos \alpha ) + c$.

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

Similar Questions

$\int \frac{d x}{x^{2}+4 x+13} = $

Find the integral of the function $\sin ^{2}(2 x+5)$.

For $-\frac{\pi}{2} < x < \frac{\pi}{2}$,evaluate the integral $\int \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) dx$ (where $C$ is a constant of integration).

$\int \frac{\text{cosec}\theta - \cot\theta}{\text{cosec}\theta + \cot\theta} \, d\theta = $

The value of $\int \frac{1}{x^2 - 5x + 8} dx$ is equal to...

Vedclass Test Series

Exam Paper Generator

Online Exam Module