Question

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

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

Answer 1

Answer: (A)

$\pi$ log2

${\int }_0^{\frac{\pi }{2}}log(\frac{sinx}{cosx}+\frac{cosx}{sinx})dx$
$={\int }_0^{\frac{\pi }{2}}log\left(\frac{1}{cosx\cdot sinx}\right)dx$
$=-{\int }_0^{\frac{\pi }{2}}log(cosx\cdot sinx)dx$
$=-\left[{\int }_0^{\frac{\pi }{2}}log\right(cosx)dx+{\int }_0^{\frac{\pi }{2}}log(sinx\left)dx\right]$
$=-[\frac{-\pi }{2}log2-\frac{\pi }{2}log2]$
$=\pi log2$