Answer
$\frac{1}{3}{\sinh ^3}x + C$
Work Step by Step
$$\eqalign{
& \int {{{\sinh }^2}x\cosh xdx} \cr
& {\text{Let }}u = \sinh x,{\text{ }}du = \cosh xdt,{\text{ substituting}} \cr
& \int {{{\sinh }^2}x\cosh xdx} = \int {{u^2}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& \int {{u^2}} du = \frac{1}{3}{u^3} + C \cr
& {\text{Write in terms of }}t,{\text{ substitute }}u = \sinh x \cr
& = \frac{1}{3}{\sinh ^3}x + C \cr} $$
