Question

If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $

• A. $\frac{\pi}{2} \log ( 1/2) $
• B. $ 1 - \frac{\pi}{2} \log ( 1/2) $
• C. $ 1 + \frac{\pi}{2} \log ( 1/2) $
• D. $\frac{\pi}{2} \log \, 2 $

Answer 1

Answer: (D)

Given, $\int\limits_{0}^{\frac{\pi}{2}} \log \cos x d x=\frac{\pi}{2} \log \frac{1}{2} \ldots$ (i)
Let $l=\int\limits_{0}^{\frac{\pi}{2}} \log \sec x d x=\int\limits_{0}^{\frac{\pi}{2}} \log \left(\frac{1}{\cos x}\right) d x$
$=\int\limits_{0}^{\frac{\pi}{2}} \log (\cos x)^{-1} d x=-\int\limits_{0}^{\frac{\pi}{2}} \log (\cos x) d x$
$=-\int\limits_{0}^{\frac{\pi}{2}} \log (\cos x) d x $
$\left[\because \log m^{-n}=-n \log m\right]$
$=-\frac{\pi}{2} \log \left(\frac{1}{2}\right) $
$\left(\because \log \left(\frac{1}{a}\right)=-\log a\right)$
$=\frac{\pi}{2} \log 2$