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Mathematics Question on Some Properties of Definite Integrals

The value of the integral ∫0π2log(tan x)dx=\int\limits_{0}^{\frac{\pi}{2}} log \left(tan\,x\right)dx=

A

0

B

1

C

π2\frac{\pi}{2}

D

π4\frac{\pi}{4}

Answer

0

Explanation

Solution

Let I = ∫0π2\int\limits_{0}^{\frac{\pi}{2}} log (tan x)dx ...(1)
Then, I = ∫0π2 log[tan(π2−x)]dx\int\limits_{0}^{\frac{\pi}{2}} \, log\bigg[ tan \bigg(\frac{\pi}{2} - x\bigg)\bigg]dx
[∵  ∫0a f(x)dx = ∫0a f(a−x)dx]\bigg[\because \, \, \int \limits_{0}^{a} \, f(x)dx \, = \, \int\limits_{0}^{a} \, f(a-x)dx\bigg]
⇒I=∫0π2 log(cotx)dx\Rightarrow I = \int\limits_{0}^{\frac{\pi}{2}} \, log(cot x)dx
⇒  I=∫0π2 log(1tan x)dx\Rightarrow \, \, I = \int\limits_{0}^{\frac{\pi}{2}} \, log\bigg(\frac{1}{tan \, x}\bigg)dx
⇒I=∫0π2 log(tanx)−1  dx=−∫0π2  log(tanx)dx\Rightarrow I = \int\limits_{0}^{\frac{\pi}{2}} \, log(tan x)^{-1} \, \, dx = -\int\limits_{0}^{\frac{\pi}{2}} \, \, log(tan x)dx
⇒\Rightarrow I = -I ⇒\Rightarrow 2I = 0 ⇒  I = 0\Rightarrow \, \, I \, = \, 0
[Using e (1)]