Question

Evaluate \( \int_{-2}^{2} x^4 (4 - x^2)^{\frac{7}{2}} \, dx \):

• A. \( 4\pi \)
• B. \( \frac{\pi}{16} \)
• C. \( 28\pi \)
• D. \( \frac{3\pi}{128} \)

Hint:

For trigonometric functions, setting the derivative equal to zero often involves solving a trigonometric equation. Remember that \(\tan(x) = -1\) has solutions at \(x = \frac{3\pi}{4} + k\pi\) for integer \(k\).

Answer 1

The Correct Option is C

To evaluate the integral \( \int_{-2}^{2} x^4 (4 - x^2)^{\frac{7}{2}} \, dx \), we can proceed with the following steps: 1. Substitution: - Let \( x = 2\sin\theta \). Then, \( dx = 2\cos\theta \, d\theta \). - When \( x = -2 \), \( \theta = -\frac{\pi}{2} \). - When \( x = 2 \), \( \theta = \frac{\pi}{2} \). - The integral becomes: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (2\sin\theta)^4 (4 - (2\sin\theta)^2)^{\frac{7}{2}} \cdot 2\cos\theta \, d\theta \] 2. Simplify the Integrand: - Simplify \( (4 - (2\sin\theta)^2)^{\frac{7}{2}} \): \[ 4 - (2\sin\theta)^2 = 4 - 4\sin^2\theta = 4(1 - \sin^2\theta) = 4\cos^2\theta \] \[ (4\cos^2\theta)^{\frac{7}{2}} = 4^{\frac{7}{2}} \cos^7\theta = 128\cos^7\theta \] - Thus, the integrand becomes: \[ (2\sin\theta)^4 \cdot 128\cos^7\theta \cdot 2\cos\theta = 2^4 \sin^4\theta \cdot 128 \cos^8\theta \cdot 2 = 2^5 \cdot 128 \sin^4\theta \cos^8\theta \] \[ = 4096 \sin^4\theta \cos^8\theta \] 3. Integrate: - The integral is now: \[ 4096 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^4\theta \cos^8\theta \, d\theta \] - Since the integrand is even, we can write: \[ 8192 \int_{0}^{\frac{\pi}{2}} \sin^4\theta \cos^8\theta \, d\theta \] - Use the beta function identity: \[ \int_{0}^{\frac{\pi}{2}} \sin^m\theta \cos^n\theta \, d\theta = \frac{1}{2} B\left(\frac{m+1}{2}, \frac{n+1}{2}\right) \] where \( B \) is the beta function. - For \( m = 4 \) and \( n = 8 \): \[ \int_{0}^{\frac{\pi}{2}} \sin^4\theta \cos^8\theta \, d\theta = \frac{1}{2} B\left(\frac{5}{2}, \frac{9}{2}\right) \] - The beta function \( B(x, y) \) is related to the gamma function: \[ B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)} \] - Calculate the gamma functions: \[ \Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4} \] \[ \Gamma\left(\frac{9}{2}\right) = \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{105\sqrt{\pi}}{16} \] \[ \Gamma\left(\frac{5}{2} + \frac{9}{2}\right) = \Gamma(7) = 6! = 720 \] - Thus: \[ B\left(\frac{5}{2}, \frac{9}{2}\right) = \frac{\frac{3\sqrt{\pi}}{4} \cdot \frac{105\sqrt{\pi}}{16}}{720} = \frac{315\pi}{46080} = \frac{7\pi}{1024} \] - Therefore: \[ \int_{0}^{\frac{\pi}{2}} \sin^4\theta \cos^8\theta \, d\theta = \frac{1}{2} \cdot \frac{7\pi}{1024} = \frac{7\pi}{2048} \] - Multiply by 8192: \[ 8192 \cdot \frac{7\pi}{2048} = 28\pi \] 4. Final Answer: - The integral evaluates to: \[ \boxed{28\pi} \] This corresponds to option (3).