int frac{1-cos x}{1+cos x} d x= . . . . . . | vedclass

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Question

(A) We know that $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.Substituting these into the integral:$\int \frac{2 \sin^2 \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$.Substituting these into the integral:$\int \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} d x = \int 	an^2 \frac{x}{2} d x$.Using the identity $	an^2 	heta = \sec^2 	heta - 1$:$\int (\sec^2 \frac{x}{2} - 1) d x$.Integrating term by term:$= \int \sec^2 \frac{x}{2} d x - \int 1 d x$.$= 2 	an \frac{x}{2} - x + C$.

$\int \frac{1-\cos x}{1+\cos x} d x=$ . . . . . . $+C$.

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