Question

By using the properties of definite integrals, evaluate the integral ${\int }_0^{\pi }\log (1+\cos x)dx$

Answer 1

Let $I={\int }_0^{\pi }\log (1+\cos x)dx$ ........... (1)
$\Rightarrow I={\int }_0^{\pi }\log (1+\cos (\pi -x\left)\right)dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\pi }\log (1-\cos x)dx$ ........... (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\pi }\left\{\log \right(1+\cos x)+\log (1-\cos x\left)\right\}dx$
$\Rightarrow 2I={\int }_0^{\pi }\log (1-{\cos }^{2}x)dx$
$\Rightarrow 2I={\int }_0^{\pi }\log {\sin }^{2}xdx$
$\Rightarrow 2I=2{\int }_0^{\pi }\log \sin xdx$
$\Rightarrow I={\int }_0^{\pi }\log \sin xdx$ ........ (3)
$\sin (\pi -x)=\sin x$
$\therefore I=2{\int }_0^{\frac{\pi }{2}}\log \sin xdx$ ............ (4)
$\Rightarrow I=2{\int }_0^{\frac{\pi }{2}}\log \sin (\frac{\pi }{2}-x)dx=2{\int }_0^{\frac{\pi }{2}}\log \cos xdx$ .............. (5)
Adding (4) and (5), we obtain
$2I=2{\int }_0^{\frac{\pi }{2}}(\log \sin x+\log \cos x)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}(\log \sin x+\log \cos x+\log 2-\log 2)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}(\log 2\sin x\cos x-\log 2)dx$
$\Rightarrow I={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$
Let $2x=t\Rightarrow 2dx=dt$
When $x=0,t=0$ and when $x=\frac{\pi }{2}$
$\therefore I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\log \sin tdt-\frac{\pi }{2}\log 2$
$\Rightarrow I=\frac{1}{2}I-\frac{\pi }{2}\log 2$
$\Rightarrow \frac{I}{2}=-\frac{\pi }{2}\log 2$
$\Rightarrow =-\pi \log 2$