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Question Number 61662 by maxmathsup by imad last updated on 06/Jun/19

$$calculate{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{cosx}{e^{\frac{1}{x}}+1}dx$$

Commented by maxmathsup by imad last updated on 07/Jun/19

$$letf\left(x\right)=\frac{cosx}{e^{\frac{1}{x}}+1}wehavethedecompositionf\left(x\right)=\frac{f\left(x\right)+f(-x)}{2}\left(even\right)+\frac{f\left(x\right)-f(-x)}{2}\left(odd\right)$$$$\Rightarrow I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{f\left(x\right)+f(-x)}{2}dx+{\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{f\left(x\right)-f(-x)}{2}dx=H+K$$$$K=0\Rightarrow I={\int }_0^{\frac{\pi }{4}}\{\frac{cos\left(x\right)}{e^{\frac{1}{x}}+1}+\frac{cosx}{e^{-\frac{1}{x}}+1}\}dx={\int }_0^{\frac{\pi }{4}}\left\{\frac{e^{-\frac{1}{x}}+1+e^{\frac{1}{x}}+1}{1+e^{\frac{1}{x}}+e^{-\frac{1}{x}}+1}\right\}cosxdx$$$$={\int }_0^{\frac{\pi }{4}}cos\left(x\right)dx={\left[sinx\right]}_0^{\frac{\pi }{4}}=\frac{\sqrt{2}}{2}\Rightarrow I=\frac{\sqrt{2}}{2}.$$

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