We are given the integral:

$\int \left( \sqrt{1 - \sin x} + \sqrt{1 + \sin x} \right) \, dx = f(x) + c$ and we are asked to find $f'(8\pi/3)$, given that $f(8\pi/3) = -2$, and the interval $5\pi/2 < x < 7\pi/2$.

Step 1: Find the derivative of $f(x)$

By the Fundamental Theorem of Calculus, we know that the derivative of the integral with respect to $x$ is simply the integrand itself. Therefore, the derivative of $f(x)$ with respect to $x$ is:

$f'(x) = \sqrt{1 - \sin x} + \sqrt{1 + \sin x}$

Thus, we need to evaluate $f'(8\pi/3)$.

Step 2: Evaluate $f'(8\pi/3)$

Substitute $x = 8\pi/3$ into the expression for $f'(x)$:

$f'(8\pi/3) = \sqrt{1 - \sin(8\pi/3)} + \sqrt{1 + \sin(8\pi/3)}$

To proceed, we need to compute $\sin(8\pi/3)$.

Step 3: Simplify $\sin(8\pi/3)$

First, express $8\pi/3$ in terms of a more familiar angle by subtracting $2\pi$:

$8\pi/3 - 2\pi = 8\pi/3 - 6\pi/3 = 2\pi/3$

Now, $\sin(8\pi/3) = \sin(2\pi/3)$. Since $\sin(2\pi/3) = \sin(\pi/3) = \frac{\sqrt{3}}{2}$, we have:

$\sin(8\pi/3) = \frac{\sqrt{3}}{2}$

Step 4: Substitute into the expression for $f'(8\pi/3)$

Now that we know $\sin(8\pi/3) = \frac{\sqrt{3}}{2}$, substitute this value into the formula for $f'(x)$:

$f'(8\pi/3) = \sqrt{1 - \frac{\sqrt{3}}{2}} + \sqrt{1 + \frac{\sqrt{3}}{2}}$

Step 5: Simplify the square roots

Let's simplify each term inside the square roots:

1. For $\sqrt{1 - \frac{\sqrt{3}}{2}}$, we calculate: $1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ Therefore: $\sqrt{1 - \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}}$

2. For $\sqrt{1 + \frac{\sqrt{3}}{2}}$, we calculate: $1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$ Therefore: $\sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{2}}$

So, the derivative $f'(8\pi/3)$ becomes:

$f'(8\pi/3) = \sqrt{\frac{2 - \sqrt{3}}{2}} + \sqrt{\frac{2 + \sqrt{3}}{2}}$

At this stage, we could approximate these values, but the final result remains in this exact form for now.

To summarize:

$f'(8\pi/3) = \sqrt{\frac{2 - \sqrt{3}}{2}} + \sqrt{\frac{2 + \sqrt{3}}{2}}$

Would you like to proceed with approximating this expression, or do you have other questions on this?

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Follow-up Questions:

1. How do we evaluate the square roots more precisely for practical use?
2. How can we simplify trigonometric integrals like this one?
3. What other methods can be used to find the derivative of integrals?
4. How do we handle integration with more complex trigonometric functions?
5. What is the relationship between the Fundamental Theorem of Calculus and the problem we solved?

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Tip: When dealing with square roots in integrals involving trigonometric functions, consider using trigonometric identities to simplify the integrand before attempting to integrate.