Question

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

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

Answer 1

Answer: (D)

$\frac{\pi }{2}\log 2$

Given : $={\int }_0^{\pi /2}\log \left(\cos x\right)dx=\frac{\pi }{2}\log \left(\frac{1}{2}\right)$ ....... $\left(i\right)$

${\int }_0^{\pi /2}\log \sec xdx$

$={\int }_0^{\pi /2}\log \left(\frac{1}{\cos x}\right)dx$
$=-{\int }_0^{\pi /2}\log \left(\cos x\right)dx$ ...... $(\log \left(\frac{1}{a}\right)=-\log a)$
$=-\frac{\pi }{2}\log \left(\frac{1}{2}\right)$ .......... From $\left(i\right)$

$=\frac{\pi }{2}\log 2$ ....... $(\because \log 1=0)$