Answer
$$\frac{{{{\cosh }^4}3 - 1}}{{12}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {{{\cosh }^3}3x} \sinh 3xdx \cr
& {\text{set }}u = \cosh 3x{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = 3\sinh 3xdx,\,\,\,\,\,\,\,\,\,\,\sinh 3xdx = \frac{1}{3}du \cr
& {\text{switch the limits of integration}} \cr
& u = \cosh 3x,\,\,\,\,x = 0 \to u = 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \to u = \cosh 3 \cr
& {\text{use the change of variable}} \cr
& \int_0^1 {{{\cosh }^3}3x} \sinh 3xdx = \int_1^{\cosh 3} {{u^3}\left( {\frac{1}{3}du} \right)} \cr
& = \frac{1}{3}\int_1^{\cosh 3} {{u^3}du} \cr
& {\text{integrate and evaluate the limits}} \cr
& = \frac{1}{3}\left( {\frac{{{u^4}}}{4}} \right)_1^{\cosh 3} \cr
& = \frac{1}{{12}}\left( {{u^4}} \right)_1^{\cosh 3} \cr
& = \frac{1}{{12}}\left( {{{\left( {\cosh 3} \right)}^4} - {{\left( 1 \right)}^4}} \right) \cr
& = \frac{{{{\cosh }^4}3 - 1}}{{12}} \cr} $$
