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

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Question

(B) Given integral: $I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$ Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get: $

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given integral: $I = \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$ Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get: $I = \int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx$ $I = \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} dx$ $I = 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$ $I = 2 \int (\cos x + \cos \alpha) dx$ Integrating with respect to $x$: $I = 2(\sin x + x \cos \alpha) + c$

$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx$ is equal to

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