Question

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

Answer 1

Let ${\int }_0^{\pi }log(1+cosx)dx$

$\Rightarrow I={\int }_0^{\pi }log\{1+cos(\pi -x\left)\right\}dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]={\int }_0^{\pi }log(1-cosx)dx[\because cos(x-x)=-cosx]={\int }_0^{\pi }log\left\{2si{n}^2\right(\frac{x}{2}\left)\right\}dx[\because 1-cosx=2si{n}^2\frac{x}{2}]={\int }_0^{pi}log2dx+2{\int }_0^{\pi }log\left(sin\frac{x}{2}\right)dx\text{In the second integral, put}(\frac{x}{2}=t\Rightarrow dx=2dt,$
$\text{and limits when}$ $x=0,t=0andwhenx=\pi ,t=\frac{\pi }{2}\therefore I=log2[x{]}_0^{\pi }+2{\int }_0^{\frac{\pi }{2}}log(sint)2dt=(log2\left)\right(\pi -0)+4(-\frac{\pi }{2}log2)(\because {\int }_0^{\frac{\pi }{2}}logsinxdx=-\frac{\pi }{2}log2)=-\pi \log 2$