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The given problem involves the evaluation of a vector-valued integral. We need to compute the integral of each component separately over the interval \([- \pi/3, \pi/3]\).

The vector function inside the integral is:

\[
(3 + 4\cos t)\mathbf{i} + (2\sin t)\mathbf{j} + (\sec^2 t)\mathbf{k}
\]

We will evaluate the integral for each component:

\[
\int_{-\pi/3}^{\pi/3} (3 + 4 \cos t) dt \quad \mathbf{i} \quad + \quad \int_{-\pi/3}^{\pi/3} (2 \sin t) dt \quad \mathbf{j} \quad + \quad \int_{-\pi/3}^{\pi/3} (\sec^2 t) dt \quad \mathbf{k}
\]

### 1. Integral for the \(\mathbf{i}\)-component:
\[
\int_{-\pi/3}^{\pi/3} (3 + 4 \cos t) dt = \int_{-\pi/3}^{\pi/3} 3 dt + \int_{-\pi/3}^{\pi/3} 4 \cos t dt
\]
The first integral is straightforward:
\[
\int_{-\pi/3}^{\pi/3} 3 dt = 3 \cdot \left[ t \right]_{-\pi/3}^{\pi/3} = 3 \cdot \left( \frac{\pi}{3} - \left( -\frac{\pi}{3} \right) \right) = 3 \cdot \frac{2\pi}{3} = 2\pi
\]

For the second part:
\[
\int_{-\pi/3}^{\pi/3} 4 \cos t dt = 4 \left[ \sin t \right]_{-\pi/3}^{\pi/3} = 4 \cdot \left( \sin\left( \frac{\pi}{3} \right) - \sin\left( -\frac{\pi}{3} \right) \right)
\]
Since \(\sin(-x) = -\sin(x)\), we have:
\[
\sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}, \quad \sin\left( -\frac{\pi}{3} \right) = -\frac{\sqrt{3}}{2}
\]
Thus:
\[
4 \left( \frac{\sqrt{3}}{2} - \left( -\frac{\sqrt{3}}{2} \right) \right) = 4 \cdot \frac{2\sqrt{3}}{2} = 4\sqrt{3}
\]

Therefore, the \(\mathbf{i}\)-component integral evaluates to:
\[
2\pi + 4\sqrt{3}
\]

### 2. Integral for the \(\mathbf{j}\)-component:
\[
\int_{-\pi/3}^{\pi/3} 2 \sin t dt = 2 \left[ -\cos t \right]_{-\pi/3}^{\pi/3} = 2 \left( -\cos\left( \frac{\pi}{3} \right) + \cos\left( -\frac{\pi}{3} \right) \right)
\]
Since \(\cos(-x) = \cos(x)\), we get:
\[
\cos\left( \frac{\pi}{3} \right) = \frac{1}{2}
\]
Thus:
\[
2 \cdot \left( -\frac{1}{2} + \frac{1}{2} \right) = 0
\]
The \(\mathbf{j}\)-component integral is 0.

### 3. Integral for the \(\mathbf{k}\)-component:
\[
\int_{-\pi/3}^{\pi/3} \sec^2 t dt = \left[ \tan t \right]_{-\pi/3}^{\pi/3} = \tan\left( \frac{\pi}{3} \right) - \tan\left( -\frac{\pi}{3} \right)
\]
Since \(\tan(-x) = -\tan(x)\), we have:
\[
\tan\left( \frac{\pi}{3} \right) = \sqrt{3}, \quad \tan\left( -\frac{\pi}{3} \right) = -\sqrt{3}
\]
Thus:
\[
\sqrt{3} - \left( -\sqrt{3} \right) = 2\sqrt{3}
\]

### Final Answer:
\[
\int_{-\pi/3}^{\pi/3} \left( (3 + 4 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + (\sec^2 t) \mathbf{k} \right) dt = (2\pi + 4\sqrt{3}) \mathbf{i} + 0 \mathbf{j} + 2\sqrt{3} \mathbf{k}
\]

So, the final vector is:
\[
(2\pi + 4\sqrt{3}) \mathbf{i} + 0 \mathbf{j} + 2\sqrt{3} \mathbf{k}
\]

Would you like any additional clarifications or details on this solution?

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Here are 5 related questions to expand your understanding:
1. How do you compute the definite integral of trigonometric functions over symmetric intervals?
2. What properties of the sine and cosine functions help simplify integrals over intervals like \([-a, a]\)?
3. What is the geometrical interpretation of vector-valued integrals?
4. How do you integrate functions involving \(\sec^2 t\), and why does it relate to the tangent function?
5. Can you apply similar integration techniques to solve problems involving parametric equations?

**Tip:** When evaluating integrals of trigonometric functions over symmetric intervals like \([-a, a]\), always check for odd and even properties, which can help simplify the calculations significantly.

Evaluate the integral of the vector-valued function [(3 + 4 cos t) i + (2 sin t) j + (sec^2 t) k] from -π/3 to π/3.

Learn how to evaluate the vector-valued integral of the function [(3 + 4 cos t) i + (2 sin t) j + (sec^2 t) k] from -π/3 to π/3. This problem covers vector calculus and trigonometric integrals, focusing on key formulas such as integrals of cosine, sine, and secant squared. Suitable for college-level calculus students.

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