Question

${\int }_0^{\pi /2}\log (\tan x+\cot x)dx$ is equal to

• A. $\frac{\pi }{2}\log 2$
• B. $-\frac{\pi }{2}\log 2$
• C. $\pi \log 2$
• D. None of these

Answer 1

The correct option is C $\pi \log 2$

Let $I={\int }_0^{\pi /2}\log (\tan x+\cot x)dx={\int }_0^{\pi /2}\log (\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})dx$
$={\int }_0^{\pi /2}\log \left(\frac{2}{\sin 2x}\right)dx=\log 2{\int }_0^{\pi /2}dx-{\int }_0^{\pi /2}\log \left(\sin 2x\right)dx$
Put $2x=z\Rightarrow dx=\frac{1}{2}dz$
$\therefore I=\frac{\pi }{2}\log 2-\frac{1}{2}{\int }_0^{\pi }\log \left(\sin z\right)dz=\frac{\pi }{2}\log 2-\frac{1}{2}.2{\int }_0^{\pi /2}\log \sin zdz$.
$=\frac{\pi }{2}\log 2+\frac{\pi }{2}\log 2=\pi \log 2$.......(as ${\int }_0^{\pi /2}\log \sin z=\frac{\pi }{2}\log 2)$