Answer
$$\frac{1}{6}\cosh 6x + C$$
Work Step by Step
$$\eqalign{
& \int {\sinh 6x} dx \cr
& {\text{Let }}u = 6x,{\text{ }}du = 6dx,{\text{ }}dx = \frac{1}{6}du \cr
& {\text{Substitute}} \cr
& \int {\sinh 6x} dx = \int {\sinh u\left( {\frac{1}{6}} \right)} du \cr
& = \frac{1}{6}\int {\sinh u} du \cr
& {\text{Using the rules of theorem 5}}{\text{.18 }}\left( {{\text{Page 385}}} \right) \cr
& = \frac{1}{6}\cosh u + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{6}\cosh 6x + C \cr} $$
