Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 305: 83

$F(6)-F(2)$

Step 1. Given $F(x)$ as an antiderivative of $f(x)$, we have $F(x)=\int f(x)dx$ and $\int_a^bf(x)dx=F(b)-F(a)$. Step 2. Letting $u=2x$, we have $du=2dx$ and the endpoints of the integral are $x=1\to u=2$ and $x=3\to u=6$ Step 3. We have $\int_1^3\frac{sin(2x)}{x}dx=\int_1^3\frac{sin(2x)}{2x}d(2x)=\int_2^6\frac{sin(u)}{u}du=F(6)-F(2)$