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

Answer: (A)

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: $9a + 8b = 2$.