\( \int _ { 0 } ^ { \pi / 3 } \frac { 1 - 8 \cos ( \theta ) ^ { 2 } } { 2 } d \theta \)

To solve the integral \( \int _ { 0 } ^ { \pi / 3 } \frac { 1 - 8 \cos ( \theta ) ^ { 2 } } { 2 } d \theta \), we first simplify the integrand: \[ \frac{1 - 8 \cos^2(\theta)}{2} = \frac{1}{2} - 4 \cos^2(\theta) \] Now, we can split the integral into two parts: \[ \int _ { 0 } ^ { \pi / 3 } \left(\frac{1}{2} - 4 \cos^2(\theta)\right) d\theta = \int _ { 0 } ^ { \pi / 3 } \frac{1}{2} d\theta - 4 \int _ { 0 } ^ { \pi / 3 } \cos^2(\theta) d\theta \] Calculating the first integral: \[ \int _ { 0 } ^ { \pi / 3 } \frac{1}{2} d\theta = \frac{1}{2} \cdot \left[\theta\right]_0^{\pi/3} = \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6} \] For the second integral, we can use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \): \[ \int _ { 0 } ^ { \pi / 3 } \cos^2(\theta) d\theta = \int _ { 0 } ^ { \pi / 3 } \frac{1 + \cos(2\theta)}{2} d\theta = \frac{1}{2} \int _ { 0 } ^ { \pi / 3 } 1 d\theta + \frac{1}{2} \int _ { 0 } ^ { \pi / 3 } \cos(2\theta) d\theta \] Calculating these: - The first part: \[ \frac{1}{2} \cdot \left[\theta\right]_0^{\pi/3} = \frac{1}{2} \cdot \frac{\pi}{3} = \frac{\pi}{6} \] - The second part: \[ \frac{1}{2} \left[\frac{\sin(2\theta)}{2}\right]_0^{\pi/3} = \frac{1}{4} \left[\sin\left(\frac{2\pi}{3}\right) - \sin(0)\right] = \frac{1}{4} \cdot \sin\left(\frac{2\pi}{3}\right) = \frac{1}{4} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8} \] Combining these: \[ \int _ { 0 } ^ { \pi / 3 } \cos^2(\theta) d\theta = \frac{\pi}{6} + \frac{\sqrt{3}}{8} \] So we have: \[ 4 \int _ { 0 } ^ { \pi / 3 } \cos^2(\theta) d\theta = 4\left(\frac{\pi}{6} + \frac{\sqrt{3}}{8}\right) = \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \] Now, let's put everything back into our integral: \[ \int _ { 0 } ^ { \pi / 3 } \left(\frac{1}{2} - 4 \cos^2(\theta)\right) d\theta = \frac{\pi}{6} - \left(\frac{2\pi}{3} + \frac{\sqrt{3}}{2}\right) \] Converting \( \frac{\pi}{6} \) to a common denominator: \[ \frac{\pi}{6} - \frac{4\pi}{6} - \frac{\sqrt{3}}{2} = -\frac{3\pi}{6} - \frac{\sqrt{3}}{2} = -\frac{\pi}{2} - \frac{\sqrt{3}}{2} \] Thus, the final answer is: \[ \int _ { 0 } ^ { \pi / 3 } \frac { 1 - 8 \cos ( \theta ) ^ { 2 } } { 2