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

##int cos(x)cosh(x)dx=[cos(x)sinh(x)+cosh(x)sin(x)]/2+C##

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Use twice

Using integration by parts once we get,

##int cos(x)cosh(x)dx=cos(x)sinh(x)+intsinh(x)sin(x)dx##

Now use integration by parts on the right hand integral

##int sinh(x)sin(x)dx=cosh(x)sin(x)-intcos(x)cosh(x)dx##

Substitute this into the top equation and rearrange to get

##2int cos(x)cosh(x)dx=cos(x)sinh(x)+cosh(x)sin(x)##

##int cos(x)cosh(x)dx=[cos(x)sinh(x)+cosh(x)sin(x)]/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.

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