Answer
$\frac{1}{2}\cos x\sinh x + \frac{1}{2}\sin x\cosh x + C $
Work Step by Step
$$\eqalign{
& \int {\cos x\cosh x} dx \cr
& {\text{Let }}u = \cos x,{\text{ }}du = - \sin xdx \cr
& dv = \cosh xdx,{\text{ }}v = \sinh x \cr
& {\text{Using the integration by parts formula}} \cr
& \int {udv} = uv - \int {vdu} \cr
& \int {\cos x\cosh x} dx = \cos x\sinh x - \int {\sinh x\left( { - \sin x} \right)dx} \cr
& \int {\cos x\cosh x} dx = \cos x\sinh x + \int {\sinh x\sin xdx} \cr
& {\text{Now}}{\text{,}} \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& dv = \sinh xdx,{\text{ }}v = \cosh x \cr
& {\text{Using the integration by parts formula}} \cr
& \int {\cos x\cosh x} dx = \cos x\sinh x + \sin x\cosh x - \int {\cosh x\cos xdx} \cr
& {\text{Add }}\int {\cosh x\cos xdx} {\text{ to both sides ofthe equation}} \cr
& 2\int {\cos x\cosh x} dx = \cos x\sinh x + \sin x\cosh x \cr
& {\text{Solve for }}\int {\cos x\cosh x} dx \cr
& \int {\cos x\cosh x} dx = \frac{1}{2}\cos x\sinh x + \frac{1}{2}\sin x\cosh x + C \cr} $$
