int frac{cos 2x - 1}{cos 2x + 1} dx = | vedclass

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(C) Let $I = \int \frac{\cos 2x - 1}{\cos 2x + 1} dx$. Using the trigonometric identities $\cos 2x = 1 - 2\sin^2 x$ and $\cos 2x = 2\cos^2 x - 1$,we h

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$x - 	an x + c$

Vedclass · Answer

(C) Let $I = \int \frac{\cos 2x - 1}{\cos 2x + 1} dx$. Using the trigonometric identities $\cos 2x = 1 - 2\sin^2 x$ and $\cos 2x = 2\cos^2 x - 1$,we have: $I = \int \frac{(1 - 2\sin^2 x) - 1}{(2\cos^2 x - 1) + 1} dx$ $I = \int \frac{-2\sin^2 x}{2\cos^2 x} dx$ $I = - \int 	an^2 x dx$ Using the identity $	an^2 x = \sec^2 x - 1$: $I = - \int (\sec^2 x - 1) dx$ $I = - \int \sec^2 x dx + \int 1 dx$ $I = - 	an x + x + c$ $I = x - 	an x + c$

$\int \frac{\cos 2x - 1}{\cos 2x + 1} dx = $

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