Question

How to solve `intcoshxsinh(sinhx)` `dx`?

Answer 1

`I = cosh(sinhx) + C`

Explanation:

We let `u =sinhx`. Then `du = coshxdx` and `dx = (du)/coshx`.

`I = int coshx sinh(u) * 1/coshx du`

`I = int sinhu du`

`I = coshu + C`

`I = cosh(sinhx) + C`

Hopefully this helps!