Question

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:

• A. 2
• B. 1
• C. 3
• D. \(\frac{3}{2}\)

Answer 1

The Correct Option is A

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.

Answer 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: \( 9a + 8b = 2 \).