If \[\int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3},\]where $a$ and $b$ are rational numbers, then $9a + 8b$ is equal to:
- 2
- 1
- 3
- \(\frac{3}{2}\)
The Correct Option is A
Solution and Explanation
To evaluate the integral:
\[ I = \int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx \]
we use the power-reduction formula:
\[ \cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x) \]
Using the formula \(\cos^2 2x = \frac{1 + \cos 4x}{2}\):
\[ \cos^4 x = \frac{1}{4}\left(1 + 2\cos 2x + \frac{1 + \cos 4x}{2}\right) = \frac{1}{4}\left(\frac{3}{2} + 2\cos 2x + \cos 4x\right) = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \]
Now, integrating term by term:
\[ I = \int_{0}^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x\right) dx \] \[ I = \frac{3}{8} \int_{0}^{\frac{\pi}{3}} dx + \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx + \frac{1}{8} \int_{0}^{\frac{\pi}{3}} \cos 4x \, dx \]
Evaluating each integral:
\[ \int_{0}^{\frac{\pi}{3}} dx = \frac{\pi}{3} \] \[ \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx = \frac{\sin 2x}{2} \bigg|_{0}^{\frac{\pi}{3}} = \frac{1}{2} \left(\sin \frac{2\pi}{3} - \sin 0\right) = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} \] \[ \int_{0}^{\frac{\pi}{3}} \cos 4x \, dx = \frac{\sin 4x}{4} \bigg|_{0}^{\frac{\pi}{3}} = \frac{1}{4} \left(\sin \frac{4\pi}{3} - \sin 0\right) = \frac{1}{4} \left(-\frac{\sqrt{3}}{2}\right) = -\frac{\sqrt{3}}{8} \]
Substituting these values:
\[ I = \frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{2} \cdot \frac{\sqrt{3}}{4} + \frac{1}{8} \cdot \left(-\frac{\sqrt{3}}{8}\right) \] \[ I = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} \]
Combining terms:
\[ I = \frac{\pi}{8} + \frac{7\sqrt{3}}{64} \]
Thus, comparing with \( I = a\pi + b\sqrt{3} \):
\[ a = \frac{1}{8}, \quad b = \frac{7}{64} \]
Calculating \( 9a + 8b \):
\[ 9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64} = \frac{9}{8} + \frac{56}{64} = \frac{9}{8} + \frac{7}{8} = 2 \]
Conclusion: \( 9a + 8b = 2 \).
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