What is the integral from 0 to 2 of 1/(e^x+e^{-x)} ?

The integral from 0 to 2 of 1/(e^x+e^{-x)} is arctan(e^2)-pi/4

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integral from 0 to 2 of 1/(e^x+e^{-x)}

\int_{0}^{2} \frac{1}{e ^{x} + e ^{- x}} d x

arctan (e^{2}) - \frac{π}{4}

Decimal

0.65088 \dots

Solution steps

\int_{0}^{2} \frac{1}{e ^{x} + e ^{- x}} d x

Apply u-substitution

: \int_{1}^{e^{2}} \frac{1}{u ^{2} + 1} d u

= \int_{1}^{e^{2}} \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= [arctan (u)]_{1}^{e^{2}}

Compute the boundaries:

arctan (e^{2}) - \frac{π}{4}

= arctan (e^{2}) - \frac{π}{4}

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What is the integral from 0 to 2 of 1/(e^x+e^{-x)} ?
The integral from 0 to 2 of 1/(e^x+e^{-x)} is arctan(e^2)-pi/4