Question

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

Answer 1

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

By adding we get $2I={\int }_0^{\pi }\log (1+cosx)dx+{\int }_0^{\pi }\log (1-cosx)dx=I={\int }_0^{\pi }\log (1-{\cos }^{2}x)dx={\int }_0^{\pi }\log \left({\sin }^{2}x\right)=2{\int }_0^{\pi }\log \left(sinx\right)$

o we get $I={\int }_0^{\pi }\log \left(sinx\right)=2{\int }_0^{\frac{\pi }{2}}\log \left(sinx\right)$

Let $J={\int }_0^{\frac{\pi }{2}}\log sinx={\int }_0^{\frac{\pi }{2}}\log cosx$

Therefore by adding , we get $2J={\int }_0^{\frac{\pi }{2}}\log sinxcosx={\int }_0^{\frac{\pi }{2}}\log 2sinx-{\int }_0^{\frac{\pi }{2}}\log 2=J-\frac{\pi }{2}\log 2$

Which implies $J=-\frac{\pi }{2}\log 2$

Therefore $I=2J=-\pi \log 2$