# How do you find the integral of (cosx)(coshx) dx?

Jun 7, 2015

$\setminus \int \cos \left(x\right) \cosh \left(x\right) \mathrm{dx} = \frac{\cos \left(x\right) \sinh \left(x\right) + \cosh \left(x\right) \sin \left(x\right)}{2} + C$

Use integration by parts twice

Using integration by parts once we get,

$\setminus \int \cos \left(x\right) \cosh \left(x\right) \mathrm{dx} = \cos \left(x\right) \sinh \left(x\right) + \setminus \int \sinh \left(x\right) \sin \left(x\right) \mathrm{dx}$

Now use integration by parts on the right hand integral

$\int \sinh \left(x\right) \sin \left(x\right) \mathrm{dx} = \cosh \left(x\right) \sin \left(x\right) - \int \cos \left(x\right) \cosh \left(x\right) \mathrm{dx}$

Substitute this into the top equation and rearrange to get

$2 \setminus \int \cos \left(x\right) \cosh \left(x\right) \mathrm{dx} = \cos \left(x\right) \sinh \left(x\right) + \cosh \left(x\right) \sin \left(x\right)$

$\setminus \int \cos \left(x\right) \cosh \left(x\right) \mathrm{dx} = \frac{\cos \left(x\right) \sinh \left(x\right) + \cosh \left(x\right) \sin \left(x\right)}{2} + C$

Where $C$ is the constant of integration. When applying integration by parts to an indefinite integral, we pick up a constant of integration. I just neglected to write it until the end.