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
Approach Solution - 1
To solve the integral \(\int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx\), we need to use trigonometric identities and integration techniques. The problem states that the integral equals \(a\pi + b\sqrt{3}\), where \(a\) and \(b\) are rational numbers, and we need to find \(9a + 8b\).
**Step 1: Use the power-reduction identity**
The power-reduction identity for \(\cos^2 x\) is:
\[\cos^2 x = \frac{1 + \cos 2x}{2}\]
Thus, \(\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2\).
**Step 2: Expand the expression**
\[ \cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4} \]
Using \(\cos^2 2x = \frac{1 + \cos 4x}{2}\), we have:
\[ \cos^4 x = \frac{1 + 2\cos 2x + \frac{1 + \cos 4x}{2}}{4} = \frac{3 + 4\cos 2x + \cos 4x}{8} \]
**Step 3: Integrate the expression**
Now, integrate each term separately from \(0\) to \(\frac{\pi}{3}\):
\[ \int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx = \int_{0}^{\frac{\pi}{3}} \frac{3}{8} \, dx + \int_{0}^{\frac{\pi}{3}} \frac{4\cos 2x}{8} \, dx + \int_{0}^{\frac{\pi}{3}} \frac{\cos 4x}{8} \, dx \]
\[ = \frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{2} \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx + \frac{1}{8} \int_{0}^{\frac{\pi}{3}} \cos 4x \, dx \]
**Step 4: Solve the integrals**
The integral \(\int \cos kx \, dx = \frac{\sin kx}{k} + C\).
Thus,
\[ \int_{0}^{\frac{\pi}{3}} \cos 2x \, dx = \left[\frac{\sin 2x}{2}\right]_{0}^{\frac{\pi}{3}} = \frac{\sin \frac{2\pi}{3}}{2} - \frac{\sin 0}{2} = \frac{\sqrt{3}}{4} \]
\[ \int_{0}^{\frac{\pi}{3}} \cos 4x \, dx = \left[\frac{\sin 4x}{4}\right]_{0}^{\frac{\pi}{3}} = \frac{\sin \frac{4\pi}{3}}{4} - \frac{\sin 0}{4} = -\frac{\sqrt{3}}{8} \]
Substitute back:
\[ \int_{0}^{\frac{\pi}{3}} \cos^4 x \, dx = \frac{\pi}{8} + \frac{1}{2} \cdot \frac{\sqrt{3}}{4} + \frac{1}{8} \left(-\frac{\sqrt{3}}{8}\right) \]
\[ = \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64} \]
Simplify:
\[ = \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64} \]
\[ = \frac{\pi}{8} + \frac{7\sqrt{3}}{64} \]
**Step 5: Find \(9a + 8b\)**
We have \(a = \frac{1}{8}\) and \(b = \frac{7}{64}\). Therefore:
\[ 9a + 8b = 9 \times \frac{1}{8} + 8 \times \frac{7}{64} \]
\[ = \frac{9}{8} + \frac{56}{64} \]
\[ = \frac{9}{8} + \frac{7}{8} = 2 \]
Thus, the value of \(9a + 8b\) is \(2\), so the correct answer is 2.
Approach Solution -2
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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