Primero integramos la función \(\frac{\cos (u)}{\sin^{2}(u)}\). Para integrar \(\frac{\cos (u)}{\sin^{2}(u)}\), usamos el método de sustitución. Sea \(v = \sin(u)\), entonces \(dv = \cos(u) du\) o \(du = \frac{dv}{\cos(u)}\). La integral se transforma en: $\int \frac{\cos (u)}{\sin^{2}(u)} \, du = \int \frac{\cos(u)}{v^{2}} \cdot \frac{dv}{\cos(u)} = \int \frac{dv}{v^{2}} = \int v^{-2} \, dv = -v^{-1} + C$ Reemplazamos \(v\) por \(\sin(u)\): $\int \frac{\cos (u)}{\sin^{2}(u)} \, du = -\frac{1}{\sin(u)} + C$ Ahora aplicamos Barrow: $ \int_{\pi / 4}^{\pi / 2} \frac{\cos (u)}{\sin^{2}(u)} \, du = \left[ -\frac{1}{\sin(u)} \right]_{\pi / 4}^{\pi / 2} $ $ \left[ -\frac{1}{\sin(u)} \right]_{\pi / 4}^{\pi / 2} = -\frac{1}{\sin(\pi / 2)} - \left(-\frac{1}{\sin(\pi / 4)}\right) $ $ = -\frac{1}{1} - \left(-\frac{1}{\frac{\sqrt{2}}{2}}\right) $ $ = -1 - \left(-\frac{2}{\sqrt{2}}\right) $ $ = -1 + \sqrt{2} $ $ = \sqrt{2} - 1 $ $ = \sqrt{2} - 1 $