Evaluate the definite integral \( \frac{5 \pi}{6} \) \( \int_{\frac{\pi}{3}} \frac{\cos x}{\sin ^{2} x} d x=\square \) (Type an exact answer.)

To evaluate the definite integral \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin ^{2} x} \, dx, \] we can use substitution and the properties of trigonometric functions. Notice that the integrand can be rewritten using the identity \( \sin^2 x = 1 - \cos^2 x \). However, a more effective approach is to recognize that \[ \frac{\cos x}{\sin^2 x} = \cot x \csc x, \] which can be integrated directly. The integral of \( \cot x \) is \( \ln |\sin x| \), so we have: \[ \int \frac{\cos x}{\sin^2 x} \, dx = -\csc x + C. \] Now we can evaluate the definite integral: \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin^2 x} \, dx = \left[-\csc x\right]_{\frac{\pi}{3}}^{\frac{5 \pi}{6}}. \] Calculating the values at the bounds, we find: \[ -\csc\left(\frac{5\pi}{6}\right) = -\frac{1}{\sin\left(\frac{5\pi}{6}\right)} = -\frac{1}{\frac{1}{2}} = -2, \] and \[ -\csc\left(\frac{\pi}{3}\right) = -\frac{1}{\sin\left(\frac{\pi}{3}\right)} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}. \] Combining these results, we get: \[ \int_{\frac{\pi}{3}}^{\frac{5 \pi}{6}} \frac{\cos x}{\sin^2 x} \, dx = -2 - \left(-\frac{2}{\sqrt{3}}\right) = -2 + \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} - 2. \] This expression matches the integration limits. Therefore, the value of the definite integral is \[ \frac{2}{\sqrt{3}} - 2. \] So, the final answer is: \[ \frac{2}{\sqrt{3}} - 2. \]