_0^ /4 1 + x 1 - x ,dx = - Vidyadip

MCQ

$\int_0^{\pi /4} {\frac{{1 + \tan x}}{{1 - \tan x}}\,dx} =$

✓
$ - \frac{1}{2}\log 2$
B
$\frac{1}{4}\log 2$
C
$\frac{1}{3}\log 2$
D
એકપણ નહીં.

✓

Answer

Correct option: A.

$ - \frac{1}{2}\log 2$

(a) $\int_0^{\pi /4} {\frac{{1 + \tan x}}{{1 - \tan x}}dx = \int_0^{\pi /4} {\tan \left( {\frac{\pi }{4} + x} \right)\,dx} } $

$ = \left[ {\log \left\{ {\sec \left( {\frac{\pi }{4} + x} \right)} \right\}} \right]_0^{ - \pi /4}$

$= - \frac{1}{2}\log 2$.

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