Question

How do you evaluate the integral `int sinhx/(1+coshx)`?

Answer 1

`int\ sinh(x)/(1+cosh(x))\ dx=ln(1+cosh(x))+C`

Explanation:

We begin by introducing a u-substitution with `u=1+cosh(x)`. The derivative of `u` is then `sinh(x)`, so we divide through by `sinh(x)` to integrate with respect to `u`:
`int\ sinh(x)/(1+cosh(x))\ dx=int\ cancel(sinh(x))/(cancel(sinh(x))*u)\ du=int\ 1/u\ du`

This integral is the common integral:
`int\ 1/t\ dt=ln|t|+C`

This makes our integral:
`ln|u|+C`

We can resubstitute to get:
`ln(1+cosh(x))+C`, which is our final answer.

We remove the absolute value from the logarithm because we note that `cosh` is positive on its domain so it's not necessary.