We are given the integral:

$\int \left( \sqrt{1 - \sin x} + \sqrt{1 + \sin x} \right) \, dx = f(x) + c$

and the condition $f(8\pi/3) = -2$, where $c$ is the constant of integration. We are asked to find $f'(8\pi/3)$.

Step 1: Differentiate the integral expression

To find $f'(x)$, we differentiate the expression for $f(x)$:

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

This is the derivative of $f(x)$ with respect to $x$, so:

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

Step 2: Evaluate $f'(x)$ at $x = \frac{8\pi}{3}$

Now, we evaluate $f'(x)$ at $x = \frac{8\pi}{3}$:

1. We first calculate $\sin\left( \frac{8\pi}{3} \right)$. To do this, we can reduce $\frac{8\pi}{3}$ within the standard interval $[0, 2\pi]$: $\frac{8\pi}{3} - 2\pi = \frac{8\pi}{3} - \frac{6\pi}{3} = \frac{2\pi}{3}$

   So, $\sin\left( \frac{8\pi}{3} \right) = \sin\left( \frac{2\pi}{3} \right)$. From the unit circle, we know: $\sin\left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}$

2. Now, we substitute $\sin\left( \frac{8\pi}{3} \right) = \frac{\sqrt{3}}{2}$ into the expression for $f'(x)$:

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

At $x = \frac{8\pi}{3}$, this becomes:

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

Now, we simplify both square roots.

• First, simplify $\sqrt{1 - \frac{\sqrt{3}}{2}}$: $1 - \frac{\sqrt{3}}{2} = \frac{2}{2} - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ So: $\sqrt{1 - \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{2}}$

• Second, simplify $\sqrt{1 + \frac{\sqrt{3}}{2}}$: $1 + \frac{\sqrt{3}}{2} = \frac{2}{2} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2}$ So: $\sqrt{1 + \frac{\sqrt{3}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{2}}$

Thus, $f'(8\pi/3)$ simplifies to:

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

This is the derivative of the function at $x = \frac{8\pi}{3}$.

---

Would you like further details on simplifying this expression or more information on the underlying concept? Let me know if you'd like to proceed!

Here are a few related questions:

1. How do you simplify square roots involving trigonometric expressions?
2. What other methods can be used to evaluate integrals involving trigonometric functions?
3. How does the fundamental theorem of calculus apply in this case?
4. What is the geometric interpretation of the function $f'(x)$?
5. How does the choice of the constant of integration affect the result of an indefinite integral?

Tip: When simplifying square roots with trigonometric functions, always try to express them in terms of simpler fractions to make your calculations easier!