Step 1: Understanding the Concept:
In calculus, definite integrals over a symmetric interval \([-a, a]\) provide an excellent opportunity to simplify calculations using the parity of the function.
An even function is one where \(f(-x) = f(x)\), implying symmetry about the y-axis.
For such functions, the area from \(-a\) to \(0\) is identical to the area from \(0\) to \(a\), effectively doubling the integral from \(0\) to \(a\).
The integrand here involves a radical term \(\sqrt{a^2 - x^2}\) raised to the 7th power.
Algebraic methods are cumbersome for such high powers, making trigonometric substitution the most efficient strategy.
By letting \(x = a \sin \theta\), the radical term transforms into a single trigonometric term via the identity \(1 - \sin^2 \theta = \cos^2 \theta\).
This transformation shifts the problem from algebraic integration to trigonometric integration over the standard interval \([0, \pi/2]\).
Finally, evaluating integrals of the form \(\int \sin^m \theta \cos^n \theta d\theta\) is best handled by Wallis' Formula, a reduction technique that avoids repetitive integration by parts.
Step 2: Key Formula or Approach:
1. Parity Property: \(\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx\) if \(f(x) = f(-x)\).
2. Substitution: Let \(x = 2 \sin \theta\), then \(dx = 2 \cos \theta d\theta\).
3. Wallis' Reduction Formula: For \(I = \int_{0}^{\pi/2} \sin^m \theta \cos^n \theta d\theta\), if \(m, n\) are even:
\[ I = \frac{(m-1)(m-3)\dots(1) \cdot (n-1)(n-3)\dots(1)}{(m+n)(m+n-2)\dots(2)} \times \frac{\pi}{2} \]
Step 3: Detailed Explanation:
Consider the function \(f(x) = x^4 (4 - x^2)^{7/2}\).
Replacing \(x\) with \(-x\): \(f(-x) = (-x)^4 (4 - (-x)^2)^{7/2} = x^4 (4 - x^2)^{7/2} = f(x)\).
Thus, the integrand is even, and the integral simplifies to:
\[ I = 2 \int_{0}^{2} x^4 (4 - x^2)^{7/2} dx \]
Now, we apply the substitution \(x = 2 \sin \theta\).
Differentiating both sides: \(dx = 2 \cos \theta d\theta\).
Adjusting limits: When \(x = 0, \sin \theta = 0 \implies \theta = 0\).
When \(x = 2, \sin \theta = 1 \implies \theta = \pi/2\).
The term \((4 - x^2)^{7/2}\) becomes \((4 - 4 \sin^2 \theta)^{7/2} = (4 \cos^2 \theta)^{7/2} = (2^2 \cos^2 \theta)^{7/2} = 2^7 \cos^7 \theta\).
Substituting these into the integral:
\[ I = 2 \int_{0}^{\pi/2} (2 \sin \theta)^4 \cdot (2^7 \cos^7 \theta) \cdot (2 \cos \theta) d\theta \]
\[ I = 2 \cdot 2^4 \cdot 2^7 \cdot 2 \int_{0}^{\pi/2} \sin^4 \theta \cos^8 \theta d\theta = 2^{13} \int_{0}^{\pi/2} \sin^4 \theta \cos^8 \theta d\theta \]
Applying Wallis' Formula for \(m=4\) and \(n=8\):
\[ \int_{0}^{\pi/2} \sin^4 \theta \cos^8 \theta d\theta = \frac{(3 \cdot 1) \cdot (7 \cdot 5 \cdot 3 \cdot 1)}{12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2} \times \frac{\pi}{2} \]
Calculating the denominator: \(12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2 = 46080\).
Calculating the numerator: \(3 \cdot 105 = 315\).
\[ \text{Integral part} = \frac{315}{46080} \cdot \frac{\pi}{2} = \frac{315}{92160} \pi = \frac{7\pi}{2048} \]
Finally, \(I = 2^{13} \cdot \frac{7\pi}{2^{11}} = 2^2 \cdot 7\pi = 4 \cdot 7\pi = 28\pi\).
Step 4: Final Answer:
The result of the integration process is \(28\pi\).