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$\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$ is equal to :
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Verified Answer
The correct answer is:
$\frac{e^2-1}{2 e}$
$I=\int_{-1}^1 \frac{\cosh x}{1+e^{2 x}} d x$
$=\int_{-1}^1 \frac{e^x+e^{-x}}{2\left(1+e^{2 x}\right)} d x\left(\because \cosh x=\frac{e^x+e^{-x}}{2}\right)$
$=\frac{1}{2} \int_{-1}^1 \frac{1+e^{2 x}}{\left(1+e^{2 x}\right) e^x} \cdot d x$
$=\frac{1}{2} \int_{-1}^1 e^{-x} d x=-\frac{1}{2}\left[e^{-x}\right]_{-1}^1$
$=-\frac{1}{2}\left(e^{-1}-e^1\right)=\frac{e^2-1}{2 e}$
$=\int_{-1}^1 \frac{e^x+e^{-x}}{2\left(1+e^{2 x}\right)} d x\left(\because \cosh x=\frac{e^x+e^{-x}}{2}\right)$
$=\frac{1}{2} \int_{-1}^1 \frac{1+e^{2 x}}{\left(1+e^{2 x}\right) e^x} \cdot d x$
$=\frac{1}{2} \int_{-1}^1 e^{-x} d x=-\frac{1}{2}\left[e^{-x}\right]_{-1}^1$
$=-\frac{1}{2}\left(e^{-1}-e^1\right)=\frac{e^2-1}{2 e}$
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