Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 296: 61

$6\ m$

Step 1. Given $v=\frac{ds}{dt}=6\ sin(2t)\ m/sec$, we have $s(t)=\int vdt=\int6sin(2t)dt=\int3sin(2t)d(2t)=-3cos(2t)+C$ where $C$ is a constant. (You can also let $u=2t, du=2dt$ to get $\int6sin(2t)dt=\int3sin(u)du=-3cos(u)+C$) Step 2. As $s(0)=0$, we have $-3cos(0)+C=0$ and thus $C=3$ Step 3. At $t=\frac{\pi}{2}$, we have $s(\frac{\pi}{2})=-3cos(2(\frac{\pi}{2}))+3=6\ m$