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

If $\int \frac{x^2+1}{x^4+1} dx = f(x) + c$,then $f(x)$ is equal to

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

If $\int \frac{(x - 1)^2}{(x^2 + 1)^2} dx = \tan^{-1} (x) + g(x) + k$,then $g(x)$ is equal to

$\int \frac{x^3-7 x+6}{x^2+3 x} \,d x=$

$\int \frac{dx}{\sin x + \sqrt{3} \cos x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module