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Evaluate:$\int \limits_0^ {\large\frac{\pi}{2}} log\tan x$

1 Answer

  • $ \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $\int \limits_0^a f(x) dx=\int \limits_0^a f(a-x) dx$
Step 1:
Given $I=\int \limits_0^{\pi/2} \log (\tan x)dx$-------(1)
By using the property $\int \limits_0^a f(x) dx=\int \limits_0^a f(a-x) dx$
$I=\int \limits_0^{\pi/2} \log (\large\frac{\pi}{2}-x)$$dx=\int \limits_0^{\pi/2} \log \cot x dx$---------(2)
Adding equ (1) and equ (2)
$2I= \int \limits_0^{\pi/2}\bigg[(\tan x)+\log \cot x\bigg]dx$
Step 2:
But we know $\log a+\log b =\log |a \times b|$
Therefore $2I= \int \limits_0^{\pi/2} \log [\tan x \times \cot x ] dx$
But $\tan x \times \cot x =\tan x \times \large\frac{1}{\tan x}=1$
Therefore $ 2I= \int \limits_0^{\pi/2} \log (1) dx$
But $\log 1=0$
Therefore $2I=0 =>I=0$
answered Apr 29, 2013 by meena.p