What is the integral from 0 to ln(2) of tanh(2x) ?

The integral from 0 to ln(2) of tanh(2x) is 1/2 ln(17/8)

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integral from 0 to ln(2) of tanh(2x)

\int_{0}^{l n (2)} tanh (2 x) d x

\frac{1}{2} ln (\frac{17}{8})

Decimal

0.37688 \dots

Solution steps

\int_{0}^{l n (2)} tanh (2 x) d x

Apply u-substitution

: \int_{0}^{2 l n (2)} tanh (u) \frac{1}{2} d u

= \int_{0}^{2 l n (2)} tanh (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 l n (2)} tanh (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int_{0}^{2 l n (2)} \frac{sinh ( u )}{cosh ( u )} d u

Apply u-substitution

: \int_{1}^{\frac{17}{8}} \frac{1}{v} d v

= \frac{1}{2} \cdot \int_{1}^{\frac{17}{8}} \frac{1}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} [ln ∣ v ∣]_{1}^{\frac{17}{8}}

Compute the boundaries:

ln (\frac{17}{8})

= \frac{1}{2} ln (\frac{17}{8})

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What is the integral from 0 to ln(2) of tanh(2x) ?
The integral from 0 to ln(2) of tanh(2x) is 1/2 ln(17/8)