Question

π/4∫ log(1+tan x)dx=π/8 log 2 ,Prove it.0

Answer 1

Step-by-step explanation:

plzz give me brainliest ans and plzzzz follow me

Answer 2

Given:

$\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$

Step-by-step explanation:

Let us consider,

$I = \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

$I = \int_{0}^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-x\right)\right) d x$

$\bold{[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x]}$

$I =\int_{0}^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan x}{1+\tan x}\right) d x$

$\bold{[\because \tan \left(\frac{\pi}{4}-x\right)= \frac{1-\tan x}{1+\tan x}\right]]}$

$I = \int_{0}^{\frac{\pi}{4}} \log \left(\frac{1+\tan x + 1-\tan x }{1+\tan x}\right) d x$

$I =\int_{0}^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan x}\right) d x$

$I =\int_{0}^{\frac{\pi}{4}} \log 2 d x-\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$

$\bold{\left[\because \log \left(\frac{a}{b}\right)=\log a-\log b\right]}$

$I =\int_{0}^{\frac{\pi}{4}} \log 2 d x-I$

$I + I=\int_{0}^{\frac{\pi}{4}} \log 2 d x$

$2 I=\int_{0}^{\frac{\pi}{4}} \log 2 d x$

$I=\log 2 \times \frac{\pi}{4} \times \frac{1}{2}$

$I =\frac{\pi}{8} \log 2$

$\therefore \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x=\frac{\pi}{8} \log 2$

Hence proved.