Aspire Faculty ID #3633 · Topic: NIMCET 2012 · Just now
NIMCET 2012

The value of integral $\displaystyle \int_{0}^{\pi/2} \log \tan x dx$ is

Solution

Let $I = \displaystyle \int_{0}^{\pi/2} \log \tan x  dx$  
Using property $\displaystyle \int_{0}^{\pi/2} f(x)dx = \int_{0}^{\pi/2} f\left(\dfrac{\pi}{2}-x\right)dx$ 
$I = \displaystyle \int_{0}^{\pi/2} \log \tan\left(\dfrac{\pi}{2}-x\right) dx$ 
$= \displaystyle \int_{0}^{\pi/2} \log \cot x  dx$ 
$= \displaystyle \int_{0}^{\pi/2} \log\left(\dfrac{1}{\tan x}\right) dx$ 
$= \displaystyle \int_{0}^{\pi/2} [-\log \tan x]  dx$ 
$I= -I$ 
$\Rightarrow I + I = 0 \Rightarrow 2I = 0 \Rightarrow I = 0$ 
Answer: $0$
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