What is the integral of cosh(mt) ?

The integral of cosh(mt) is 1/m sinh(mt)+C

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integral of cosh(mt)

\int cosh (m t) d t

\frac{1}{m} sinh (m t) + C

Solution steps

\int cosh (m t) d t

Apply u-substitution

: \int \frac{cosh ( u )}{m} d u

= \int \frac{cosh ( u )}{m} d u

Take the constant out:

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

= \frac{1}{m} \cdot \int cosh (u) d u

Use the common integral:

\int cosh (u) d u = sinh (u)

= \frac{1}{m} sinh (u)

Substitute back

u = m t

= \frac{1}{m} sinh (m t)

Add a constant to the solution

= \frac{1}{m} sinh (m t) + C

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