Question

\(\int_{-2\pi}^{2\pi} \sin^2(2x) \cos^4(2x) \, dx =\)

• A. \(\frac{3\pi}{64}\)
• B. \(\frac{9\pi}{64}\)
• C. \(\frac{9\pi}{35}\)
• D. \(\frac{9\pi}{280}\)

Hint:

For even trigonometric integrands, use \(\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx\). Apply identities like \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\) or Wallis’s integrals for \(\int \sin^m x \cos^n x \, dx\).

Answer 1

The Correct Option is A

The integrand \(\sin^2(2x) \cos^4(2x)\) is even, so: \[ I = \int_{-2\pi}^{2\pi} \sin^2(2x) \cos^4(2x) \, dx = 2 \int_0^{2\pi} \sin^2(2x) \cos^4(2x) \, dx \] Substitute \( u = 2x \), \( du = 2 \, dx \), \( dx = \frac{du}{2} \), with limits from \( u = 0 \) to \( u = 4\pi \): \[ I = 2 \cdot \frac{1}{2} \int_0^{4\pi} \sin^2 u \cos^4 u \, du = \int_0^{4\pi} \sin^2 u \cos^4 u \, du \] Use trigonometric identities: \(\sin^2 u = \frac{1 - \cos 2u}{2}\), \(\cos^2 u = \frac{1 + \cos 2u}{2}\), so \(\cos^4 u = \left( \frac{1 + \cos 2u}{2} \right)^2 = \frac{1 + 2 \cos 2u + \cos^2 2u}{4}\). \[ \sin^2 u \cos^4 u = \frac{1 - \cos 2u}{2} \cdot \frac{1 + 2 \cos 2u + \cos^2 2u}{4} = \frac{1}{8} (1 - \cos 2u)(1 + 2 \cos 2u + \cos^2 2u) \] Expand: \[ (1 - \cos 2u)(1 + 2 \cos 2u + \cos^2 2u) = 1 + 2 \cos 2u + \cos^2 2u - \cos 2u - 2 \cos^2 2u - \cos^3 2u \] \[ = 1 + \cos 2u - \cos^2 2u - \cos^3 2u \] \[ I = \frac{1}{8} \int_0^{4\pi} (1 + \cos 2u - \cos^2 2u - \cos^3 2u) \, du \] Evaluate each term: \[ \int_0^{4\pi} 1 \, du = [u]_0^{4\pi} = 4\pi \] \[ \int_0^{4\pi} \cos 2u \, du = \left[ \frac{\sin 2u}{2} \right]_0^{4\pi} = 0 \] \[ \cos^2 2u = \frac{1 + \cos 4u}{2}, \quad \int_0^{4\pi} \cos^2 2u \, du = \int_0^{4\pi} \frac{1 + \cos 4u}{2} \, du = \left[ \frac{u}{2} + \frac{\sin 4u}{8} \right]_0^{4\pi} = \frac{4\pi}{2} = 2\pi \] For \(\cos^3 2u\), use \(\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta\): \[ \cos^3 2u = \frac{\cos 6u + 3 \cos 2u}{4}, \quad \int_0^{4\pi} \cos^3 2u \, du = \int_0^{4\pi} \frac{\cos 6u + 3 \cos 2u}{4} \, du = 0 \] \[ I = \frac{1}{8} \left( 4\pi + 0 - 2\pi - 0 \right) = \frac{1}{8} \cdot 2\pi = \frac{\pi}{4} \] Alternatively, use Wallis’s integrals over one period \([0, \frac{\pi}{2}]\): \[ \int_0^{\frac{\pi}{2}} \sin^m u \cos^n u \, du = \frac{(m-1)(m-3)\cdots (2 \text{ or } 1) \cdot (n-1)(n-3)\cdots (2 \text{ or } 1)}{(m+n)(m+n-2)\cdots (2 \text{ or } 1)} \cdot \frac{\pi}{2}, \quad \text{for } m, n \text{ even} \] For \( m = 2 \), \( n = 4 \): \[ \int_0^{\frac{\pi}{2}} \sin^2 u \cos^4 u \, du = \frac{1 \cdot (3 \cdot 1)}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{48} \cdot \frac{\pi}{2} = \frac{\pi}{32} \] The period of \(\sin^2(2x) \cos^4(2x)\) is \(\frac{\pi}{2}\), so from \( 0 \) to \( 4\pi \), there are \(\frac{4\pi}{\pi/2} = 8\) periods: \[ \int_0^{4\pi} \sin^2 u \cos^4 u \, du = 8 \cdot \frac{\pi}{32} = \frac{\pi}{4} \] Thus, \( I = \frac{\pi}{4} \). Compare with options: \(\frac{3\pi}{64} \approx 0.147\), \(\frac{9\pi}{64} \approx 0.442\), \(\frac{9\pi}{35} \approx 0.808\), \(\frac{9\pi}{280} \approx 0.101\), but \(\frac{\pi}{4} \approx 0.785\). None match, confirming “None of the given options is correct.”