Answer
\[ = \sqrt 2 - 1\]
Work Step by Step
\[\begin{gathered}
\int_0^{\frac{\pi }{4}} {\frac{{\sin x}}{{{{\cos }^2}x}}} \,\,\,dx \hfill \\
\hfill \\
set\,\,\,u = \cos x\,\,\,\,{\text{which implies that}}\,\,\,du = - \sin xdx \hfill \\
\hfill \\
{\text{Changing limits of integration}} \hfill \\
\hfill \\
x = 0\,\,\,\,implies\,\,u = 1 \hfill \\
x = \frac{\pi }{4}\,\,\,implies\,\,u = \frac{{\sqrt 2 }}{2} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_1^{\frac{{\sqrt 2 }}{2}} {\frac{{du}}{{{u^2}}}} \hfill \\
\hfill \\
integrate\,\, \hfill \\
\hfill \\
= \,\,\left[ {\frac{1}{u}} \right]_1^{\frac{{\sqrt 2 }}{2}} \hfill \\
\hfill \\
Fundamental\,\,theorem \hfill \\
\hfill \\
= \frac{1}{{\frac{{\sqrt 2 }}{2}}} - \frac{1}{1} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \sqrt 2 - 1 \hfill \\
\end{gathered} \]
