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

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(D) We have the integral $I = \int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx}$.Using the trigonometric identities $\cos x = 1 - 2\sin^2 \frac{x}{2}$ and

Vedclass · Accepted Answer

$x - 2	an \frac{x}{2} + c$

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(D) We have the integral $I = \int {\frac{{\cos x - 1}}{{\cos x + 1}}\,dx}$.Using the trigonometric identities $\cos x = 1 - 2\sin^2 \frac{x}{2}$ and $\cos x = 2\cos^2 \frac{x}{2} - 1$,we get:$I = \int {\frac{{(1 - 2\sin^2 \frac{x}{2}) - 1}}{{(2\cos^2 \frac{x}{2} - 1) + 1}}\,dx}$$I = \int {\frac{{ - 2\sin^2 \frac{x}{2}}}{{2\cos^2 \frac{x}{2}}}\,dx} = - \int {	an^2 \frac{x}{2}\,dx}$Using the identity $	an^2 	heta = \sec^2 	heta - 1$,we have:$I = - \int {(\sec^2 \frac{x}{2} - 1)\,dx} = \int {(1 - \sec^2 \frac{x}{2})\,dx}$Integrating term by term,we get:$I = x - 2	an \frac{x}{2} + c$.

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

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