Question

Evaluate: ${\int }_0^{\pi /2}\log \left(\tan x\right)dx$

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $\pi$

Answer 1

Answer: (C)

$0$

Let, $I={\int }_0^{\pi /2}\log \left(\tan x\right)dx$

Using the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$, we get:-

$I={\int }_0^{\pi /2}\log \left(\tan (\frac{\pi }{2}-x\right))dx$

$⟹I={\int }_0^{\pi /2}\log \left(\cot x\right)dx$

$={\int }_0^{\pi /2}\log \frac{1}{\tan x}dx$

$=-{\int }_0^{\pi /2}\log \left(\tan x\right)dx$

$I=-I$

$⟹I=0$

Hence,answer is option-(C).