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

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(C) हम जानते हैं कि $	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}$.अतः,

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) हम जानते हैं कि $	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}$.अतः,समाकलन $\int 	an \left( \frac{\pi}{4} - x ight) \, dx$ हो जाता है।सूत्र $\int 	an 	heta \, d	heta = \log |\sec 	heta| + c$ या $-\log |\cos 	heta| + c$ का उपयोग करते हुए:$\int 	an \left( \frac{\pi}{4} - x ight) \, dx = -\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| \cdot (-1) + c = \log \left| \sec \left( \frac{\pi}{4} - x ight) ight| + c$.वैकल्पिक रूप से,$\cos \left( \frac{\pi}{4} - x ight) = \sin \left( \frac{\pi}{2} - \left( \frac{\pi}{4} - x ight) ight) = \sin \left( \frac{\pi}{4} + x ight)$ का उपयोग करने पर,हमें प्राप्त होता है:$-\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| + c = \log \left| \sin \left( \frac{\pi}{4} + x ight) ight| + c$.

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

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