What is the integral from 2 to 3 of 1/(e^t-e^{-t)} ?

The integral from 2 to 3 of 1/(e^t-e^{-t)} is 1/2 ln((e^5+e^3-e^2-1)/(e^5-e^3+e^2-1))

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integral from 2 to 3 of 1/(e^t-e^{-t)}

\int_{2}^{3} \frac{1}{e ^{t} - e ^{- t}} d t

\frac{1}{2} ln (\frac{e ^{5} + e ^{3} - e ^{2} - 1}{e ^{5} - e ^{3} + e ^{2} - 1})

Decimal

0.08634 \dots

Solution steps

\int_{2}^{3} \frac{1}{e ^{t} - e ^{- t}} d t

Rewrite using trig identities

= \int_{2}^{3} \frac{1}{2 sinh ( t )} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{2}^{3} \frac{1}{sinh ( t )} d t

Use the Hyperbolic identity:

\frac{1}{sinh ( x )} = csch (x)

= \frac{1}{2} \cdot \int_{2}^{3} csch (t) d t

Use the common integral:

\int csch (t) d t = ln (tanh (\frac{t}{2}))

= \frac{1}{2} [ln (tanh (\frac{t}{2}))]_{2}^{3}

Compute the boundaries:

ln (\frac{e ^{3} - 1}{e ^{3} + 1}) - ln (\frac{e ^{2} - 1}{e ^{2} + 1})

= \frac{1}{2} (ln (\frac{e ^{3} - 1}{e ^{3} + 1}) - ln (\frac{e ^{2} - 1}{e ^{2} + 1}))

Simplify

= \frac{1}{2} ln (\frac{e ^{5} + e ^{3} - e ^{2} - 1}{e ^{5} - e ^{3} + e ^{2} - 1})

What is the integral from 2 to 3 of 1/(e^t-e^{-t)} ?
The integral from 2 to 3 of 1/(e^t-e^{-t)} is 1/2 ln((e^5+e^3-e^2-1)/(e^5-e^3+e^2-1))