What is \int_{0}^{\pi}(\sin^{4}x+\cos^{4}x)dx equal to?
Consider the following for the next two (02) items that follow: Let I=\int_{-2\pi}^{2\pi}\frac{\sin^{4}x+\cos^{4}x}{1+3^{x}}dx
- A. \frac{3\pi}{8}
- B. \frac{3\pi}{4} ✓
- C. \frac{3\pi}{2}
- D. 3\pi
Correct Answer: B. \frac{3\pi}{4}
Explanation
We can rewrite \sin^4 x + \cos^4 x as 1 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x) = 1 - \frac{1}{4}(1 - \cos 4x) = \frac{3}{4} + \frac{1}{4}\cos 4x. Integrating this from 0 to \pi: \int_0^\pi (\frac{3}{4} + \frac{1}{4}\cos 4x) dx = [\frac{3}{4}x + \frac{1}{16}\sin 4x]_0^\pi = \frac{3\pi}{4} + 0 = \frac{3\pi}{4}.
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