Question

${\int }_0^{\pi /2}\log \left(\frac{cosx}{sinx}\right)dx$ is equal to

• A. $\pi$/2
• B. $\pi$/4
• C. $\pi$
• D. 2$\pi$
• E. 0

Answer 1

Answer: (E)

Let $I={\int }_0^{\pi /2}\log \left(\frac{\cos x}{\sin x}\right)dx={\int }_0^{\pi /2}\log (\cot x)dx...(i)$
Then, $I={\int }_0^{\pi /2}\log \cot \left(\frac{\pi }{2}-x\right)dx$
$[\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx]....(ii)$
$={\int }_0^{\pi /2}\log \tan xdx$
On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi /2}(\log \cot x+\log \tan x)dx$
$\Rightarrow 2l={\int }_0^{\pi /2}0dx=0$
$\Rightarrow l=0$