Question

The value of ${\int }_0^{\pi }\log (1+\cos x)dx$ is

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

Answer 1

The correct option is B $\pi \log \frac{1}{2}$

Let

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

$I={\int }_0^{\pi }\log \{1+\cos (\pi -x\left)\right\}dx$

$={\int }_0^{\pi }\log (1-\cos x)dx$ ...... (ii)

On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\pi }\left\{\log \right(1+\cos x)+\log (1-\cos x\left)\right\}dx$

$I=\frac{1}{2}{\int }_0^{\pi }\log (1-{\cos }^{2}x)dx$

$=\frac{1}{2}{\int }_0^{\pi }\log {\sin }^{2}xdx$

$={\int }_0^{\pi }\log \sin xdx$

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

$\left(\because {\int }_0^{2a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx,iff(2a-x)=f\left(s\right)\right)$

$=2(-\frac{\pi }{2}\log 2)$

$\left(\because {\int }_0^{\pi /2}\log \sin xdx=-\frac{\pi }{2}\log 2\right)$

$=\pi \log \frac{1}{2}$