Question

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

• A. $\frac{\pi }{2}$
• B. $0$
• C. $1$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (B)

$0$

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

Take $x$ as $\frac{\pi }{2}-x$, we get

$I={\int }_{\frac{\pi }{2}}^0\log (\tan \left(\frac{\pi }{2}\right)-x)(-dx)$

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

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

Now, $2I={\int }_0^{\frac{\pi }{2}}\log \left(\tan x\right)+\log \left(\cot x\right)dx$

$={\int }_0^{\frac{\pi }{2}}\log (\tan x.\cot x)dx$

$={\int }_0^{\frac{\pi }{2}}\log \left(1\right)dx$

$=0$

Hence $I=0$.