Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 25

$\frac{1}{2}sin^6(\frac{x}{3})+C$

Let $u=sin(\frac{x}{3})=>du=\frac{1}{3}cos(\frac{x}{3})dx=>3du=cos(\frac{x}{3})dx$ =$\int sin^5(\frac{x}{3})cos(\frac{x}{3})$ =$\int u^5(3du)$ =$3(\frac{1}{6}u^6)+C$ =$\frac{1}{2}sin^6(\frac{x}{3})+C$