Question

If ${\int }_0^{\pi /2}\log \cos xdx=\frac{\pi }{2}\log \left(\frac{1}{2}\right)$ then ${\int }_0^{\pi /2}\log \sec xdx=$

• 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, $\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log \cos xdx=\frac{\pi }{2}\log \frac{1}{2}\dots$ (i)
Let $l=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log \sec xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log \left(\frac{1}{\cos x}\right)dx$
$=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log (\cos x{)}^{-1}dx=-\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log (\cos x)dx$
$=-\underset{0}{\overset{\frac{\pi }{2}}{\int }}\log (\cos x)dx$
$\left[\because \log m^{-n}=-n\log m\right]$
$=-\frac{\pi }{2}\log \left(\frac{1}{2}\right)$
$\left(\because \log \left(\frac{1}{a}\right)=-\log a\right)$
$=\frac{\pi }{2}\log 2$