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

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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 = $

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