How do you integrate #int (cosx)(coshx)# using integration by parts?

1 Answer
Apr 11, 2018

below

Explanation:

#I = int \ cosx \ coshx \ dx#

#= int \ cosx \ d(sinh x)#

#= cosx sinh x - int \ d( cosx) \ sinh x#

#= cosx sinh x + int sin x \ sinh x \ dx#

#= cosx sinh x + int sin x \ d(cosh x)#

#= cosx sinh x + (sin x cosh x - int \ d(sin x )\ cosh x)#

#= cosx sinh x + sin x cosh x - I#

#implies I = 1/2 ( cosx sinh x + sin x cosh x) + C#