Question

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

• A. $\frac{\pi }{2}\log 2$
• B. $-\pi \log 2$
• C. $\pi \log 2$
• D. None of these

Answer 1

The correct option is B $-\pi \log 2$

Given,

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

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

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

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

adding (1) and (2)

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

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

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

$\therefore I={\int }_0^{\pi }\log \sin xdx$ ......... (3)
$\Rightarrow I=2{\int }_0^{\frac{\pi }{2}}\log \left(\sin x\right)dx$ ......(4)

$\Rightarrow I=2{\int }_0^{\frac{\pi }{2}}\log \left(\cos x\right)dx$ ........(5)
adding (4) and (5)
$\Rightarrow 2I=2{\int }_0^{\frac{\pi }{2}}\log (\sin x\cdot \cos x)dx$

$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \left(\frac{2\sin x\cdot \cos x}{2}\right)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \sin \left(2x\right)dx-{\int }_0^{\frac{\pi }{2}}\log 2dx$
put 2x=t
$I={\int }_0^{\pi }\log \sin t\cdot \frac{dt}{2}-\frac{\pi }{2}\log 2$
$I=\frac{1}{2}{\int }_0^{\pi }\log \sin x\cdot dx-\frac{\pi }{2}\log 2$
using (3)
$I=\frac{1}{2}I-\frac{\pi }{2}\log 2$
$\therefore I={\int }_0^{\pi }\log (1+\cos x)dx={\int }_0^{\pi }\log \sin xdx=-\pi \log 2$