Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.8 Hyperbolic Functions And Hanging Cables - Exercises Set 6.8 - Page 481: 29

$$ = \frac{{{{\sinh }^7}x}}{7} + C$$

$$\eqalign{ & \int {{{\sinh }^6}x\cosh xdx} \cr & {\text{substitute }}u = \sinh x,{\text{ }}du = \cosh xdx \cr & = \int {{{\sinh }^6}x\cosh xdx} = \int {{u^6}du} \cr & {\text{find the antiderivarive by the power rule}} \cr & = \frac{{{u^7}}}{7} + C \cr & {\text{write in terms of }}x \cr & = \frac{{{{\left( {\sinh x} \right)}^7}}}{7} + C \cr & = \frac{{{{\sinh }^7}x}}{7} + C \cr} $$