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.)

Evaluate the integral by following steps: - step0: Evaluate using substitution: \(\int_{\frac{\pi }{3}}^{\frac{5\pi }{6}} \frac{\cos\left(x\right)}{\sin^{2}\left(x\right)} dx\) - step1: Evaluate the integral: \(\int \frac{\cos\left(x\right)}{\sin^{2}\left(x\right)} dx\) - step2: Use the substitution \(dx=\frac{1}{\cos\left(x\right)} dt\) to transform the integral\(:\) \(\int \frac{\cos\left(x\right)}{\sin^{2}\left(x\right)}\times \frac{1}{\cos\left(x\right)} dt\) - step3: Simplify: \(\int \frac{1}{\sin^{2}\left(x\right)} dt\) - step4: Use the substitution \(t=\sin\left(x\right)\) to transform the integral\(:\) \(\int \frac{1}{t^{2}} dt\) - step5: Evaluate the integral: \(\frac{t^{-2+1}}{-2+1}\) - step6: Add the numbers: \(\frac{t^{-1}}{-2+1}\) - step7: Add the numbers: \(\frac{t^{-1}}{-1}\) - step8: Divide the terms: \(-t^{-1}\) - step9: Express with a positive exponent: \(-\frac{1}{t}\) - step10: Substitute back: \(-\frac{1}{\sin\left(x\right)}\) - step11: Return the limits: \(\left(-\frac{1}{\sin\left(x\right)}\right)\bigg |_{\frac{\pi }{3}}^{\frac{5\pi }{6}}\) - step12: Calculate the value: \(\frac{2\sqrt{3}-6}{3}\) To evaluate the definite integral \[ \int_{\frac{\pi}{3}}^{\frac{5\pi}{6}} \frac{\cos x}{\sin^{2} x} \, dx, \] we can use the substitution method. 1. **Substitution**: Let \( u = \sin x \). Then, \( du = \cos x \, dx \). 2. **Change of limits**: When \( x = \frac{\pi}{3} \), \( u = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). When \( x = \frac{5\pi}{6} \), \( u = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \). Now, the integral becomes: \[ \int_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}} \frac{1}{u^2} \, du. \] 3. **Integrate**: The integral of \( \frac{1}{u^2} \) is \( -\frac{1}{u} \). 4. **Evaluate the definite integral**: \[ -\frac{1}{u} \bigg|_{\frac{\sqrt{3}}{2}}^{\frac{1}{2}} = -\left(-2 + \frac{2}{\sqrt{3}}\right) = 2 - \frac{2}{\sqrt{3}}. \] 5. **Final result**: The exact answer is \[ \frac{2\sqrt{3} - 6}{3}. \] Thus, \[ \int_{\frac{\pi}{3}}^{\frac{5\pi}{6}} \frac{\cos x}{\sin^{2} x} \, dx = \frac{2\sqrt{3} - 6}{3}. \]