The given expression is:

$2 \int_{0}^{\frac{\pi}{3}} \sqrt{2 + 2 \cos \theta} \, d\theta$

Step 1: Simplifying the Square Root Expression

We can factor the term inside the square root as follows:

$\sqrt{2 + 2 \cos \theta} = \sqrt{2(1 + \cos \theta)} = \sqrt{2} \cdot \sqrt{1 + \cos \theta}$

Thus, the integral becomes:

$2 \int_{0}^{\frac{\pi}{3}} \sqrt{2} \cdot \sqrt{1 + \cos \theta} \, d\theta = 2\sqrt{2} \int_{0}^{\frac{\pi}{3}} \sqrt{1 + \cos \theta} \, d\theta$

Step 2: Using a Trigonometric Identity

We use the identity:

$1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right)$

So, the integral becomes:

$2\sqrt{2} \int_{0}^{\frac{\pi}{3}} \sqrt{2 \cos^2\left(\frac{\theta}{2}\right)} \, d\theta = 2\sqrt{2} \int_{0}^{\frac{\pi}{3}} \sqrt{2} \cos\left(\frac{\theta}{2}\right) \, d\theta$

This simplifies to:

$4 \int_{0}^{\frac{\pi}{3}} \cos\left(\frac{\theta}{2}\right) \, d\theta$

Step 3: Changing the Variable of Integration

Let $u = \frac{\theta}{2}$, so $du = \frac{1}{2} d\theta$, and the limits change accordingly:

• When $\theta = 0$, $u = 0$.
• When $\theta = \frac{\pi}{3}$, $u = \frac{\pi}{6}$.

The integral becomes:

$4 \cdot 2 \int_{0}^{\frac{\pi}{6}} \cos u \, du = 8 \int_{0}^{\frac{\pi}{6}} \cos u \, du$

Step 4: Integrating

The integral of $\cos u$ is $\sin u$. Thus, we have:

$8 [\sin u]_{0}^{\frac{\pi}{6}} = 8 \left( \sin\left(\frac{\pi}{6}\right) - \sin(0) \right) = 8 \left(\frac{1}{2}\right) = 4$

Final Answer:

$\boxed{4}$

If you have any questions or would like more details, feel free to ask!

Here are 5 questions related to this problem:

1. What other trigonometric identities could be useful in similar integrals?
2. How can we compute definite integrals involving trigonometric functions more generally?
3. What are the key steps in handling integrals with square roots involving cosine or sine?
4. What role does variable substitution play in simplifying integrals?
5. Can this method be extended to other angles or limits of integration?

Tip: When dealing with integrals involving trigonometric functions, using appropriate identities like double-angle or half-angle formulas can greatly simplify the problem.