Indefinite integral via u-substitution

Let $u = \sinh(3x)$. Then $du = 3\cosh(3x)\,dx$, so $\cosh(3x)\,dx = \dfrac{du}{3}$. Substituting into the integral gives $\int \sinh(3x)\cosh(3x)\,dx = \int u \cdot \dfrac{du}{3} = \dfrac{1}{3}\int u\, du$.

Evaluating this basic power-rule integral gives $\dfrac{1}{3}\cdot \dfrac{u^2}{2} + C = \dfrac{u^2}{6} + C$. Substituting back $u = \sinh(3x)$ yields the final answer $\dfrac{\sinh^2(3x)}{6} + C$.