Answer
\[ = \sqrt 2 - 1\]
Work Step by Step
\[\begin{gathered}
\int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\frac{{\cos x}}{{{{\sin }^2}x}}\,dx} \hfill \\
\hfill \\
set\,\,u = \sin x\,\,\,then\,\,\,du = \cos xdx \hfill \\
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x = \frac{\pi }{4}\,\,\,\,implies\,\,u = \sin \,\left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2} \hfill \\
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x = \frac{\pi }{2}\,\,\,\,\,\,implies\,\,u = \sin \,\left( {\frac{\pi }{2}} \right) = 1 \hfill \\
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apply\,\,the\,\,\,substitution \hfill \\
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\int_{\frac{{\sqrt 2 }}{2}}^1 {\frac{{du}}{{{u^2}}}} \hfill \\
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integrate \hfill \\
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= \,\,\left[ { - \frac{1}{u}} \right]_{\frac{{\sqrt 2 }}{2}}^1\, = \,\,\left[ {\frac{1}{u}} \right]_1^{\frac{{\sqrt 2 }}{2}} \hfill \\
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Fundamental\,\,theorem \hfill \\
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= \frac{1}{{\frac{{\sqrt 2 }}{2}}} - \frac{1}{1} \hfill \\
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Simplify \hfill \\
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= \sqrt 2 - 1 \hfill \\
\hfill \\
\end{gathered} \]