What is the integral of cosh(5t-10) ?

The integral of cosh(5t-10) is 1/5 sinh(5t-10)+C

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integral of cosh(5t-10)

\int cosh (5 t - 10) d t

\frac{1}{5} sinh (5 t - 10) + C

Solution steps

\int cosh (5 t - 10) d t

Apply u-substitution

: \int cosh (u) \frac{1}{5} d u

= \int cosh (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int cosh (u) d u

Use the common integral:

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

= \frac{1}{5} sinh (u)

Substitute back

u = 5 t - 10

= \frac{1}{5} sinh (5 t - 10)

Add a constant to the solution

= \frac{1}{5} sinh (5 t - 10) + C

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What is the integral of cosh(5t-10) ?
The integral of cosh(5t-10) is 1/5 sinh(5t-10)+C