\( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x}{\sin ^{2} x} d x=\square \) (Type an exact answer.)

To evaluate the integral \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x}{\sin^2 x} \, dx \), we can use a substitution. Let \( u = \sin x \). Then, the derivative is \( du = \cos x \, dx \). The limits of integration also change: when \( x = \frac{\pi}{4} \), \( u = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \); when \( x = \frac{\pi}{2} \), \( u = \sin\left(\frac{\pi}{2}\right) = 1 \). Now we rewrite the integral: \[ \int_{\frac{\sqrt{2}}{2}}^{1} \frac{1}{u^2} \, du \] The integral of \( \frac{1}{u^2} \) is \( -\frac{1}{u} \). Thus, we have: \[ \left[ -\frac{1}{u} \right]_{\frac{\sqrt{2}}{2}}^{1} \] Now calculating the definite integral: \[ -\frac{1}{1} - \left( -\frac{1}{\frac{\sqrt{2}}{2}} \right) = -1 + \frac{2}{\sqrt{2}} = -1 + \sqrt{2} \] So, the value of the integral is: \[ \sqrt{2} - 1 \] Therefore, the answer is: \[ \sqrt{2} - 1 \]