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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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} = $

Similar Questions

Integrate the function $\sin (ax+b) \cos (ax+b)$.

$\int {\frac{{1 + {{\tan }^2}x}}{{1 - {{\tan }^2}x}}\,dx} $ is equal to

If $g\left(\frac{t+1}{2 t+1}\right)=t+1$,then $\int g(x) d x=$

$\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}} dx = $ (where $C$ is a constant of integration.)

$\int \tan^4 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module