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

To solve the given expression \( \frac { 1 } { 2 } \int _ { 0 } ^ { \pi / 3 } 1 d \theta - \int _ { 0 } ^ { \pi / 3 } 8 \cos \theta ^ { 2 } d \theta \), we can break it down step by step. First, we calculate the first integral: \[ \frac{1}{2} \int_{0}^{\pi/3} 1 \, d\theta = \frac{1}{2} \left[ \theta \right]_{0}^{\pi/3} = \frac{1}{2} \left( \frac{\pi}{3} - 0 \right) = \frac{\pi}{6}. \] Next, we compute the second integral involving \( \cos^2 \theta \). We can use the identity \( \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \): \[ \int_{0}^{\pi/3} 8 \cos^2 \theta \, d\theta = \int_{0}^{\pi/3} 8 \cdot \frac{1 + \cos(2\theta)}{2} \, d\theta = 4 \int_{0}^{\pi/3} (1 + \cos(2\theta)) \, d\theta. \] Now, we compute this integral: \[ 4 \left( \int_{0}^{\pi/3} 1 \, d\theta + \int_{0}^{\pi/3} \cos(2\theta) \, d\theta \right). \] The first part is \( \int_{0}^{\pi/3} 1 \, d\theta = \frac{\pi}{3} \). For the second part, \[ \int \cos(2\theta) \, d\theta = \frac{1}{2} \sin(2\theta), \] we evaluate from 0 to \( \frac{\pi}{3} \): \[ \left[ \frac{1}{2} \sin(2\theta) \right]_{0}^{\pi/3} = \frac{1}{2} \sin\left(\frac{2\pi}{3}\right) - 0 = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}. \] Putting it all together: \[ \int_{0}^{\pi/3} \cos(2\theta) d\theta = 0 + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{4}. \] Thus, \[ 4 \left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right) = \frac{4\pi}{3} + \sqrt{3}. \] Finally, we substitute back into the original expression: \[ \frac{\pi}{6} - \left(\frac{4\pi}{3} + \sqrt{3}\right). \] To combine the terms, we find a common denominator: \[ \frac{\pi}{6} - \frac{8\pi}{6} - \sqrt{3} = -\frac{7\pi}{6} - \sqrt{3}. \] Therefore, the final result is: \[ -\frac{7\pi}{6} - \sqrt{3}. \]