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

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(C) We know that $	an \left( \frac{\pi}{4} - x ight) = \frac{	an(\pi/4) - 	an x}{1 + 	an(\pi/4) 	an x} = \frac{1 - 	an x}{1 + 	an x}$.Therefo

Vedclass · Accepted Answer

$\log \sin \left( \frac{\pi}{4} + x \right) + c$

Vedclass · Answer

(C) We know that $	an \left( \frac{\pi}{4} - x ight) = \frac{	an(\pi/4) - 	an x}{1 + 	an(\pi/4) 	an x} = \frac{1 - 	an x}{1 + 	an x}$.Therefore,the integral becomes $\int 	an \left( \frac{\pi}{4} - x ight) \, dx$.Using the formula $\int 	an 	heta \, d	heta = \log |\sec 	heta| + c$ or $-\log |\cos 	heta| + c$,we get:$\int 	an \left( \frac{\pi}{4} - x ight) \, dx = -\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| \cdot (-1) + c = \log \left| \sec \left( \frac{\pi}{4} - x ight) ight| + c$.Alternatively,using $\cos \left( \frac{\pi}{4} - x ight) = \sin \left( \frac{\pi}{2} - \left( \frac{\pi}{4} - x ight) ight) = \sin \left( \frac{\pi}{4} + x ight)$,we have:$-\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| + c = \log \left| \frac{1}{\cos \left( \frac{\pi}{4} - x ight)} ight| + c = \log \left| \sec \left( \frac{\pi}{4} - x ight) ight| + c$.Wait,checking the options: $\log \sin \left( \frac{\pi}{4} + x ight) + c$ is equivalent to $-\log \cos \left( \frac{\pi}{4} - x ight) + c$. Since $\cos(\frac{\pi}{4}-x) = \sin(\frac{\pi}{2} - (\frac{\pi}{4}-x)) = \sin(\frac{\pi}{4}+x)$,the result is $\log \sin(\frac{\pi}{4}+x) + c$.

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

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