Evaluate the integral. \[ \int_{\pi / 2}^{0} \cos (\theta) d \theta \]

To evaluate the integral \[ \int_{\pi / 2}^{0} \cos (\theta) \, d \theta, \] we can first switch the limits of integration to change the sign of the integral. Therefore, we have: \[ \int_{\pi / 2}^{0} \cos (\theta) \, d \theta = -\int_{0}^{\pi / 2} \cos (\theta) \, d \theta. \] Now, we will compute the integral \(\int_{0}^{\pi / 2} \cos (\theta) \, d \theta\). The antiderivative of \(\cos(\theta)\) is \(\sin(\theta)\). So we evaluate: \[ \int \cos(\theta) \, d\theta = \sin(\theta) + C. \] Now we evaluate the definite integral: \[ \int_{0}^{\pi / 2} \cos(\theta) \, d\theta = \left[ \sin(\theta) \right]_{0}^{\pi / 2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1. \] Thus, we find: \[ -\int_{0}^{\pi / 2} \cos(\theta) \, d\theta = -1. \] Combining all of these steps, we obtain the final evaluation of the original integral: \[ \int_{\pi / 2}^{0} \cos (\theta) \, d \theta = -1. \] Therefore, the solution is: \[ \boxed{-1}. \]