Question

${\int }_0^{\pi }\log (1+\cos x)dx=?$

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

Answer 1

Answer: (A), (B)

The correct options are
A $\pi \log \frac{1}{2}.$
B $-\pi \log 2.$

$I={\int }_0^{\pi }\log (1+\cos x)dx$
$I={\int }_0^{\pi }\log [1+\cos (\pi -x)]dx,$ by above Prop.
or $I={\int }_0^{\pi }\log (1-\cos x)dx$
$\therefore 2I={\int }_0^{\pi }[\log (1+\cos x)+\log (1-\cos x)]dx$
$={\int }_0^{\pi }\log (1-{\cos }^{2}x)dx={\int }_0^{\pi }\log {\sin }^{2}xdx$
$=2{\int }_0^{\pi }\log \sin xdx$
$=2.2{\int }_0^{\pi /2}\log \sin xdx$
$=4.\frac{\pi }{2}.\log \frac{1}{2}.$
$\therefore 2I=2\pi \log \frac{1}{2}$ or $I=\pi \log \frac{1}{2}.$