How do you find the integral of ##(cosx)(coshx) dx##?
February 20th, 2023
##int cos(x)cosh(x)dx=[cos(x)sinh(x)+cosh(x)sin(x)]/2+C##
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.