_ ^ 1 - x 1 + x ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{1 - \tan x}}{{1 + \tan x}}\;dx = } $

A
$\log \sec \left( {\frac{\pi }{4} - x} \right) + c$
B
$\log \cos \left( {\frac{\pi }{4} + x} \right) + c$
✓
$\log \sin \left( {\frac{\pi }{4} + x} \right) + c$
D
None of these

✓

Answer

Correct option: C.

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

c
(c)$\int_{}^{} {\frac{{1 - \tan x}}{{1 + \tan x}}\,dx} = \int_{}^{} {\tan \left( {\frac{\pi }{4} - x} \right)} \,dx$
$ = \int_{}^{} {\frac{{\sin \left( {\frac{\pi }{4} - x} \right)}}{{\cos \left( {\frac{\pi }{4} - x} \right)}}} \,dx = \log \cos \left( {\frac{\pi }{4} - x} \right) + c$
$ = \log \sin \left( {\frac{\pi }{4} + x} \right) + c$.

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