To solve this problem, we need to evaluate the integral

$\int_{1}^{-1} \vec{r}(t) \, dt$

where

$\vec{r}(t) = \tan(t) \, \vec{i} - \sin^3(t) \cos^2(t) \, \vec{j} - 8t \, \vec{k}.$

This integral involves vector components, so we’ll evaluate each component separately.

Step-by-Step Solution:

1. Split the Vector Integral by Components: Since $\vec{r}(t) = \tan(t) \, \vec{i} - \sin^3(t) \cos^2(t) \, \vec{j} - 8t \, \vec{k}$, the integral can be separated into components: $\int_{1}^{-1} \vec{r}(t) \, dt = \int_{1}^{-1} \tan(t) \, dt \, \vec{i} - \int_{1}^{-1} \sin^3(t) \cos^2(t) \, dt \, \vec{j} - \int_{1}^{-1} 8t \, dt \, \vec{k}.$

2. Change the Limits of Integration: Since the limits are from $1$ to $-1$, we can reverse them to go from $-1$ to $1$ and multiply the result by $-1$: $-\int_{-1}^{1} \tan(t) \, dt \, \vec{i} + \int_{-1}^{1} \sin^3(t) \cos^2(t) \, dt \, \vec{j} + \int_{-1}^{1} 8t \, dt \, \vec{k}.$

3. Evaluate Each Integral:

   • For $\int_{-1}^{1} \tan(t) \, dt$: The function $\tan(t)$ is an odd function (i.e., $\tan(-t) = -\tan(t)$), and since the interval $[-1, 1]$ is symmetric around $0$, this integral evaluates to $0$.

     $\int_{-1}^{1} \tan(t) \, dt = 0$

   • For $\int_{-1}^{1} \sin^3(t) \cos^2(t) \, dt$: Similarly, $\sin^3(t) \cos^2(t)$ is also an odd function because $\sin^3(-t) \cos^2(-t) = -\sin^3(t) \cos^2(t)$. Therefore, this integral also evaluates to $0$.

     $\int_{-1}^{1} \sin^3(t) \cos^2(t) \, dt = 0$

   • For $\int_{-1}^{1} 8t \, dt$: The function $8t$ is an odd function, so this integral also evaluates to $0$.

     $\int_{-1}^{1} 8t \, dt = 0$

4. Combine the Results: Since each component integral evaluates to $0$, the entire vector integral evaluates to

   $\int_{1}^{-1} \vec{r}(t) \, dt = 0 \, \vec{i} + 0 \, \vec{j} + 0 \, \vec{k} = \vec{0}.$

Final Answer:

$\int_{1}^{-1} \vec{r}(t) \, dt = \vec{0}.$

Would you like more details on specific steps or have any questions about this process?

Further Related Questions:

1. What are even and odd functions, and why do they impact integrals over symmetric intervals?
2. How does the reversal of limits affect the result of an integral?
3. Could a non-zero result ever occur if only one component was an odd function?
4. What would the solution look like if any of the vector components were even functions?
5. How would you integrate $\tan(t)$, $\sin^3(t) \cos^2(t)$, or $8t$ over an interval that isn’t symmetric?

Quick Tip:

When integrating vector functions, always check the symmetry of each component function. This can simplify the integration process, especially for odd functions over symmetric intervals.