Step 1: Understanding the Concept:
Integration often requires algebraic intuition to transform a complex-looking fraction into a simpler polynomial or radical form.
The term inside the parenthesis \(\frac{4}{x^2} - 1\) is a quotient that can be combined under a common denominator.
When this common denominator \(x^2\) is raised to the power of \(5/2\), it results in \(x^5\), which can then be divided out by the \(x^8\) term outside.
This simplification reveals an integrand of the form \(x^m (a^2 - x^2)^n\), which is the standard signal for trigonometric substitution.
Trigonometric substitution works by replacing the algebraic variable with a function that exploits the Pythagorean identity, converting radical constraints into periodic functions.
This transformation is specifically designed to handle the square root of quadratic differences.
Step 2: Key Formula or Approach:
1. Simplify \(\left(\frac{4-x^2}{x^2}\right)^{5/2} = \frac{(4-x^2)^{5/2}}{(x^2)^{5/2}} = \frac{(4-x^2)^{5/2}}{x^5}\).
2. Substitution: \(x = 2 \sin \theta \implies dx = 2 \cos \theta d\theta\).
3. Wallis' Formula: For \(\int_{0}^{\pi/2} \sin^m \theta \cos^n \theta d\theta\), if one exponent is odd, the factor is \(1\) (no \(\pi/2\)).
Step 3: Detailed Explanation:
First, we rewrite the integrand to consolidate powers of \(x\):
\[ x^8 \left( \frac{4-x^2}{x^2} \right)^{5/2} = x^8 \cdot \frac{(4-x^2)^{5/2}}{x^5} = x^3 (4-x^2)^{5/2} \]
Now, the integral to solve is \(I = \int_{0}^{2} x^3 (4-x^2)^{5/2} dx\).
Let \(x = 2 \sin \theta\). Then \(dx = 2 \cos \theta d\theta\).
Changing the limits: When \(x = 0, \theta = 0\); when \(x = 2, \theta = \pi/2\).
Substituting the values:
\[ I = \int_{0}^{\pi/2} (2 \sin \theta)^3 \cdot (4 - 4 \sin^2 \theta)^{5/2} \cdot (2 \cos \theta) d\theta \]
\[ I = \int_{0}^{\pi/2} 8 \sin^3 \theta \cdot (4 \cos^2 \theta)^{5/2} \cdot 2 \cos \theta d\theta \]
\[ I = 8 \cdot 2^5 \cdot 2 \int_{0}^{\pi/2} \sin^3 \theta \cdot \cos^5 \theta \cdot \cos \theta d\theta = 2^9 \int_{0}^{\pi/2} \sin^3 \theta \cos^6 \theta d\theta \]
Using Wallis' formula for \(m=3\) and \(n=6\):
\[ \text{Value} = \frac{(2) \cdot (5 \cdot 3 \cdot 1)}{(3+6)(3+6-2)(3+6-4)\dots} = \frac{2 \cdot 15}{9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} = \frac{30}{945} = \frac{2}{63} \]
Finally, multiply by the constant: \(I = 2^9 \cdot \frac{2}{63} = \frac{2^{10}}{63}\).
Step 4: Final Answer:
The final result of the definite integral is \(\frac{2^{10}}{63}\).