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

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Question

(C) આપણે જાણીએ છીએ કે $	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}$.તે

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $	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}$.તેથી,સંકલન $\int 	an \left( \frac{\pi}{4} - x ight) \, dx$ બને છે.સૂત્ર $\int 	an 	heta \, d	heta = \log |\sec 	heta| + c$ અથવા $-\log |\cos 	heta| + c$ નો ઉપયોગ કરતા:$\int 	an \left( \frac{\pi}{4} - x ight) \, dx = -\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| \cdot (-1) + c = \log \left| \sec \left( \frac{\pi}{4} - x ight) ight| + c$.વૈકલ્પિક રીતે,$\cos \left( \frac{\pi}{4} - x ight) = \sin \left( \frac{\pi}{2} - \left( \frac{\pi}{4} - x ight) ight) = \sin \left( \frac{\pi}{4} + x ight)$ નો ઉપયોગ કરતા,આપણને મળે છે:$-\log \left| \cos \left( \frac{\pi}{4} - x ight) ight| + c = \log \left| \sin \left( \frac{\pi}{4} + x ight) ight| + c$.

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

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