## Calculus (3rd Edition)

$$\frac{1}{9}\cos(4-9x)+C$$
Finding the general antiderivative of $f$: $\int{f(x)}dx$ $=\int\sin(4-9x)dx$ Given $$\int\sin({kx})dx=-\frac{1}{k}\cos({kx})+C$$ $\int\sin(4-9x)dx$ $=-\frac{1}{-9}\cos(4-9x)$ $=\frac{1}{9}\cos(4-9x)+C$