MCQ
$\int_0^{\pi / 2} \log (\tan x) d x=$
  • A
    $\frac{\pi}{2} \log _e 2$
  • B
    $-\frac{\pi}{2} \log _{ e } 2$
  • C
    $\pi \log _e 2$
  • $0$

Answer

Correct option: D.
$0$
(D)
$\int_0^{\frac{\pi}{2}} \log \tan x d x=\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin x}{\cos x}\right) d x$
$\begin{array}{l}=\int_0^{\frac{\pi}{2}} \log \sin x d x-\int_0^{\frac{\pi}{2}} \log \cos x d x \\ =\int_0^{\frac{\pi}{2}} \log \sin x d x-\int_0^{\frac{\pi}{2}} \log \sin x d x\end{array}$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$=0$

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