Question

The value of the integral ${\int }_0^{\frac{\pi }{2}}log|tanx+cotx|dx$ is

• A. $\pi log2$
• B. $-\pi log2$
• C. $\frac{\pi }{2}log2$
• D. $0$

Answer 1

The correct option is A $\pi log2$

Given, ${\int }_0^{\frac{\pi }{2}}\log |\tan x+\cot x|dx={\int }_0^{\frac{\pi }{2}}\log \left|\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right|dx$
$={\int }_0^{\frac{\pi }{2}}\log \left|\frac{1}{\sin x\cos x}\right|dx=-{\int }_0^{\frac{\pi }{2}}\log \left|\sin x\right|dx-{\int }_0^{\frac{\pi }{2}}\log \left|\cos x\right|dx$
$=-(-\frac{\pi }{2}\log 2)-(-\frac{\pi }{2}\log 2)$

$=\pi \log 2$