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

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Question

(C) We know that $\cos 2	heta = 2\cos^2 	heta - 1$. Substituting this into the integral,we get: $\int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{

Vedclass · Accepted Answer

$2 \sin x + 2x \cos \alpha + c$,where $c$ is the constant of integration.

Vedclass · Answer

(C) We know that $\cos 2	heta = 2\cos^2 	heta - 1$. Substituting this into the integral,we get: $\int \frac{(2\cos^2 x - 1) - (2\cos^2 \alpha - 1)}{\cos x - \cos \alpha} dx$ $= \int \frac{2\cos^2 x - 2\cos^2 \alpha}{\cos x - \cos \alpha} dx$ $= 2 \int \frac{(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$ $= 2 \int (\cos x + \cos \alpha) dx$ $= 2 (\sin x + x \cos \alpha) + c$ $= 2 \sin x + 2x \cos \alpha + c$.

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

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