Question

for $mathbf{r}(t)=e^{-3t}mathbf{i}+sin(5t)mathbf{j}+cos(6t)mathbf{k}$, determine $mathbf{r}(t)$ and $intmathbf{r}(t)dt=mathbf{i}+mathbf{j}+mathbf{k}+mathbf{c}$ with $mathbf{c}$ a constant vector.

Explanation:

Step1: Integrate each component separately

We know that if $\mathbf{r}(t)=e^{-3t}\mathbf{i}+\sin(5t)\mathbf{j}+\cos(6t)\mathbf{k}$, then $\int\mathbf{r}(t)dt=\int e^{-3t}dt\mathbf{i}+\int\sin(5t)dt\mathbf{j}+\int\cos(6t)dt\mathbf{k}$.
For $\int e^{-3t}dt$, let $u = - 3t$, $du=-3dt$, so $\int e^{-3t}dt=-\frac{1}{3}e^{-3t}+C_1$.
For $\int\sin(5t)dt$, let $u = 5t$, $du = 5dt$, so $\int\sin(5t)dt=-\frac{1}{5}\cos(5t)+C_2$.
For $\int\cos(6t)dt$, let $u = 6t$, $du=6dt$, so $\int\cos(6t)dt=\frac{1}{6}\sin(6t)+C_3$.

Step2: Combine the results

$\int\mathbf{r}(t)dt=-\frac{1}{3}e^{-3t}\mathbf{i}-\frac{1}{5}\cos(5t)\mathbf{j}+\frac{1}{6}\sin(6t)\mathbf{k}+\mathbf{C}$, where $\mathbf{C}=C_1\mathbf{i}+C_2\mathbf{j}+C_3\mathbf{k}$ is a constant vector.

Answer:

$-\frac{1}{3}e^{-3t}\mathbf{i}-\frac{1}{5}\cos(5t)\mathbf{j}+\frac{1}{6}\sin(6t)\mathbf{k}+\mathbf{C}$