int_{-1}^1 frac{cosh x}{1+e^{2 x}} d x is equ | vedclass

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Question

(C) Let $I = \int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$. Since $\cosh x = \frac{e^x + e^{-x}}{2}$,we substitute this into the integral: $I = \int_{-1}

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$\frac{e^2-1}{2 e}$

Vedclass · Answer

(C) Let $I = \int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$. Since $\cosh x = \frac{e^x + e^{-x}}{2}$,we substitute this into the integral: $I = \int_{-1}^1 \frac{e^x + e^{-x}}{2(1 + e^{2x})} d x$. We can rewrite the numerator as $\frac{e^x(1 + e^{-2x})}{2(1 + e^{2x})} = \frac{e^x(1 + \frac{1}{e^{2x}})}{2(1 + e^{2x})} = \frac{e^x(\frac{e^{2x} + 1}{e^{2x}})}{2(1 + e^{2x})} = \frac{e^x}{2e^{2x}} = \frac{1}{2e^x} = \frac{1}{2}e^{-x}$. Thus,$I = \int_{-1}^1 \frac{1}{2}e^{-x} d x$. $I = \frac{1}{2} \left[ -e^{-x} \right]_{-1}^1$. $I = -\frac{1}{2} (e^{-1} - e^1) = \frac{1}{2} (e^1 - e^{-1}) = \frac{e - \frac{1}{e}}{2} = \frac{e^2 - 1}{2e}$.

$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :

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