Integral exercise asked by Adnan Y. I. Ajez in mathematics teacher group

Exercise:

Integrate, $\int{x\cos x\cosh x\,dx}$

Solution: Let $u=x\cos x\,\,\And \,\,dv=\cosh x\,dx$ $\Leftrightarrow du=\left( \cos x-x\sin x \right)\,dx\,\,\And \,\,\,v=\sinh x$

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

                            $=x\cos x\sinh x-\int{\sinh x\cos x}\,dx+\int{x\sin x\sinh x}\,dx$   (*)

Let $U=x\sin x\,\,\,\,\And \,\,dV=\sinh x\,dx\Leftrightarrow dU=\left( \sin x+x\cos x \right)\,dx\,\,\,\And \,\,V=\cosh x$

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

                          $=x\sin x\cosh x-\int{\cosh x\sin x\,dx}-\int{x\cos x\cosh x}\,dx$  (**)

Let $u=\sin x\,\,\,\And \,\,\,dv=\cosh x\,dx\Leftrightarrow du=\cos x\,dx\,\,\And \,v=\sinh x$

So $\int{\sin x\cosh x\,dx}=\sin x\sinh x-\int{\sinh x\cos x\,dx}$   (***)

$2\int{x\cos x\cosh x\,dx}=x\cos x\sinh x-\int{\cos x\sinh x\,dx}\,+x\sin x\cosh x-\int{\sin x\cosh x\,dx}$

$=x\cos x\sinh x+x\sin x\cosh x-\int{\cos x\sinh x\,dx-\sin x\sinh x+\int{\cos x\sinh x\,dx}}$

Thus $\int{x\cos x\cosh x\,dx}=\frac{1}{2}x\cos x\sinh x+\frac{1}{2}x\sin x\cosh x-\frac{1}{2}\sin x\sinh x+c$