Question

The value of ${\int }_0^{\pi /2}\log |\tan x+\cot x|dx$ is

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

Answer 1

Answer: (A)

$\underset{0}{\overset{\pi /2}{\int }}log\left|tanx+cotx\right|dx$
$=\underset{0}{\overset{\pi /2}{\int }}log\left|\frac{si{n}^{2}x+co{s}^{2}x}{cosxsinx}\right|dx$
$=\underset{0}{\overset{\pi /2}{\int }}log\left(\frac{1}{sinxcosx}\right)dx$
[$\because$ $sinx,cosxare+Ve$ in the first quadrant]
$=\underset{0}{\overset{\pi /2}{\int }}log1dx-\underset{0}{\overset{\pi /2}{\int }}logsinxdx-\underset{0}{\overset{\pi /2}{\int }}logcosxdx$
$=0-\left(-\frac{\pi }{2}log2\right)-\left(-\frac{\pi }{2}log2\right)$
$=\frac{\pi }{2}log2+\frac{\pi }{2}log2=\pi log2$