Question

${\int }_0^{\frac{\pi }{2}}log(tanx+cotx)dx=$

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

Answer 1

The correct option is A $\pi log2$

${\int }_0^{\frac{\pi }{2}}log(tanx+cotx)dx={\int }_0^{\frac{\pi }{2}}log\left[\frac{2}{sin2x}\right]dx$
$={\int }_0^{\frac{\pi }{2}}(log2-logsin2x)dx=log2\left(\frac{\pi }{2}\right)+\left(\frac{\pi }{2}\right)log2=\pi log2$