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}

Vedclass · Accepted Answer

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

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: $I = \int_{-1}^1 \frac{e^x + e^{-x}}{2(1 + e^{2x})} d x$. Factor out $e^x$ from the numerator term $e^x + e^{-x}$: $e^x + e^{-x} = e^x(1 + e^{-2x}) = e^x \left(1 + \frac{1}{e^{2x}}\right) = e^x \left(\frac{e^{2x} + 1}{e^{2x}}\right) = \frac{1 + e^{2x}}{e^x}$. Substituting this back into the integral: $I = \frac{1}{2} \int_{-1}^1 \frac{1 + e^{2x}}{e^x(1 + e^{2x})} d x$. Canceling the term $(1 + e^{2x})$: $I = \frac{1}{2} \int_{-1}^1 e^{-x} d x$. Evaluating the integral: $I = \frac{1}{2} [-e^{-x}]_{-1}^1 = -\frac{1}{2} [e^{-1} - e^1] = \frac{1}{2} (e^1 - e^{-1}) = \frac{e^2 - 1}{2e}$.

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

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