Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 7

$-\frac{1}{3}(cos(3x))+c$

$u=3x$, therefore $du=3dx$. Let's rewrite the integral! $\int (sin3x)dx$ We do the substitution! $\int (sin3x)dx=\int (sinu)\frac{du}{3}=\frac{1}{3}\int (sinu)du=\frac{1}{3}(-cosu)+c$ We re-do the substitution! $-\frac{1}{3}(cosu)+c=-\frac{1}{3}(cos(3x))+c$