Question

$\underset{-1/2}{\overset{1/2}{\int }}\cos x\log \left(\frac{1+x}{1-x}\right)dx$

Answer 1

$$\begin{gathered} {\int }_{\frac{-1}{2}}^{\frac{1}{2}}\cos x\log \left(\frac{1+x}{1-x}\right)dx \\ \mathrm{Let}f\left(x\right)=\cos x\log \left(\frac{1+x}{1-x}\right) \\ \mathrm{Consider}f\left(-x\right)=\cos \left(-x\right)\log \left(\frac{1-x}{1+x}\right) \\ =\cos x\left\{-\log \left(\frac{1+x}{1-x}\right)\right\}=-\cos x\log \left(\frac{1+x}{1-x}\right)=-f\left(x\right) \\ \mathrm{Thus}f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function} \\ \mathrm{Therefore}, \\ {\int }_{\frac{-1}{2}}^{\frac{1}{2}}\cos x\log \left(\frac{1+x}{1-x}\right)dx=0 \end{gathered}$$