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The odd part of this peculiar integral $$f(x)=\int_{0}^x\frac{\cos(z)}{e^{\frac{1}{z}}+1}dz$$ seems to be equal to half the sine function, that is $$f(x)-f(-x)=\sin(x)$$ I have only observed this graphically through Desmos, but could anyone explain why this is true?

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$$ f(-x) = \int_0^{-x} \dfrac{\cos(z)\; dz}{\exp(1/z)+1} = -\int_0^x \dfrac{\cos(z)\; dz}{\exp(-1/z)+1} $$ Now note that $$ \dfrac{1}{\exp(1/z)+1} + \frac{1}{\exp(-1/z)+1} = \frac{1}{\exp(1/z)+1} + \frac{\exp(1/z)}{1+\exp(1/z)} = 1 $$

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$$(f(x)-f(-x))'=\frac{\cos x}{e^{1/x}+1}+\frac{\cos x}{e^{-1/x}+1}=\cos x.$$

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