Use a change of variables or the table of general integration formulas to evaluate the following definite integral. \( \int_{-2 \pi}^{5 \pi} \sin \frac{\theta}{6} d \theta \) \( \int_{-2 \pi}^{5 \pi} \sin \frac{\theta}{6} d \theta=\square \) (Type an exact answer, using radicals view the table of general integration formulas. Ty

To evaluate the definite integral \( \int_{-2 \pi}^{5 \pi} \sin \frac{\theta}{6} \, d\theta \), we can use the following change of variables. Let \( u = \frac{\theta}{6} \), which implies \( d\theta = 6 \, du \). The limits of integration also change: when \( \theta = -2\pi \), \( u = -\frac{\pi}{3} \), and when \( \theta = 5 \pi \), \( u = \frac{5\pi}{6} \). Now, the integral becomes: \[ \int_{-\frac{\pi}{3}}^{\frac{5\pi}{6}} 6 \sin u \, du \] Evaluating this, we find: \[ 6 \left[-\cos u\right]_{-\frac{\pi}{3}}^{\frac{5\pi}{6}} = 6 \left(-\cos\left(\frac{5\pi}{6}\right) + \cos\left(-\frac{\pi}{3}\right)\right) \] Calculating the cosine values, \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) and \(\cos\left(-\frac{\pi}{3}\right) = \frac{1}{2}\), we get: \[ 6 \left(-\left(-\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\right) = 6 \left(\frac{\sqrt{3}}{2} + \frac{1}{2}\right) = 6 \cdot \frac{\sqrt{3}+1}{2} = 3(\sqrt{3}+1) \] Therefore, the definite integral evaluates to: \[ \int_{-2 \pi}^{5 \pi} \sin \frac{\theta}{6} \, d\theta = 3(\sqrt{3}+1) \]