int frac{cos 2x - cos 2theta}{cos x - cos th | vedclass

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Question

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

Vedclass · Accepted Answer

$ 2(\sin x + x \cos 	heta) + C $

Vedclass · Answer

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

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

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