We are tasked with evaluating the integral

$\int_0^2 \sqrt{x(4 - x)} \, dx$

using the substitution $x = 4 \sin^2 \theta$.

Step-by-Step Solution:

1. Substitution: Let $x = 4 \sin^2 \theta$, then $dx = 8 \sin \theta \cos \theta \, d\theta = 4 \sin(2\theta) \, d\theta.$

2. Change of limits:

   • When $x = 0$, $4 \sin^2 \theta = 0 \Rightarrow \sin \theta = 0 \Rightarrow \theta = 0$.
   • When $x = 2$, $4 \sin^2 \theta = 2 \Rightarrow \sin^2 \theta = \frac{1}{2} \Rightarrow \sin \theta = \frac{\sqrt{2}}{2} \Rightarrow \theta = \frac{\pi}{4}$.

   So the new limits of integration are from $\theta = 0$ to $\theta = \frac{\pi}{4}$.

3. Simplifying the integrand: Substituting $x = 4 \sin^2 \theta$, we get: $4 - x = 4 - 4 \sin^2 \theta = 4 \cos^2 \theta.$ Therefore, $\sqrt{x(4 - x)} = \sqrt{4 \sin^2 \theta \cdot 4 \cos^2 \theta} = 4 \sin \theta \cos \theta = 2 \sin(2\theta).$

4. Rewrite the integral: The original integral becomes: $\int_0^2 \sqrt{x(4 - x)} \, dx = \int_0^{\frac{\pi}{4}} 2 \sin(2\theta) \cdot 4 \sin(2\theta) \, d\theta.$ This simplifies to: $8 \int_0^{\frac{\pi}{4}} \sin^2(2\theta) \, d\theta.$

5. Using a trigonometric identity: Recall that $\sin^2 \theta = \frac{1}{2}(1 - \cos(2\theta))$. Thus, $\sin^2(2\theta) = \frac{1}{2}(1 - \cos(4\theta)).$ The integral becomes: $8 \int_0^{\frac{\pi}{4}} \frac{1}{2} \left( 1 - \cos(4\theta) \right) d\theta = 4 \int_0^{\frac{\pi}{4}} \left( 1 - \cos(4\theta) \right) d\theta.$

6. Integrating: The integral is: $4 \left[ \theta - \frac{\sin(4\theta)}{4} \right]_0^{\frac{\pi}{4}}.$ At $\theta = \frac{\pi}{4}$: $\theta = \frac{\pi}{4}, \quad \sin(4\theta) = \sin(\pi) = 0.$ At $\theta = 0$: $\theta = 0, \quad \sin(0) = 0.$ So the expression simplifies to: $4 \left( \frac{\pi}{4} - 0 \right) = \pi.$

Final Answer:

The value of the integral is $\pi$.

Would you like any further details or have any questions?

Here are 5 related questions:

1. How does the substitution $x = 4 \sin^2 \theta$ simplify the integral?
2. Can you show another example using a trigonometric substitution?
3. What is the geometric interpretation of this integral?
4. How do you handle integrals involving products of square roots and polynomials?
5. What are the typical cases where $\sin^2 \theta$ substitution is useful?

Tip: When solving integrals, substitutions involving trigonometric identities often help simplify square roots or complex polynomials.