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Question Number 95199 by Fikret last updated on 23/May/20

$${\int }_{-\pi }^{\pi }cosxln\frac{1+x}{1-x}dx=?$$

Answered by mr W last updated on 23/May/20

$$f\left(x\right)=\cos x\ln \frac{1+x}{1-x}$$$$f(-x)=\cos (-x)\ln \frac{1+(-x)}{1-(-x)}$$$$=\cos (-x)\ln \frac{1-x}{1+x}$$$$=\cos x\ln {\left(\frac{1+x}{1-x}\right)}^{-1}$$$$=-\cos x\ln \frac{1+x}{1-x}$$$$=-f\left(x\right)$$$$domainoffunctionf\left(x\right)is(-1,+1)$$$$therefore{\int }_{-\pi }^{\pi }f\left(x\right)dxisnotdefined.$$$$thereforeichangeditto$$$${\int }_{-\pi /4}^{\pi /4}f\left(x\right)dx=?.$$$$$$$${\int }_{-\pi /4}^{\pi /4}f\left(x\right)dx={\int }_{-\pi /4}^{0}f\left(x\right)dx+{\int }_0^{\pi /4}f\left(x\right)dx$$$$={\int }_{-\pi /4}^{0}f(-x)d(-x)+{\int }_0^{\pi /4}f\left(x\right)dx$$$$={\int }_{\pi /4}^{0}f\left(x\right)dx+{\int }_0^{\pi /4}f\left(x\right)dx$$$$=-{\int }_0^{\pi /4}f\left(x\right)dx+{\int }_0^{\pi /4}f\left(x\right)dx$$$$=0$$

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