Đáp án:
$\begin{array}{l}
\int\limits_0^{\pi /4} {\frac{{\cos u}}{{1 - {{\sin }^2}u}}du} \\
= \int\limits_0^{\pi /4} {\frac{1}{{\left( {1 - \sin u} \right)\left( {1 + \sin u} \right)}}d\sin u} \\
= \int\limits_0^{1/\sqrt 2 } {\frac{1}{{\left( {1 - t} \right)\left( {1 + t} \right)}}dt} \\
= \int\limits_0^{1/\sqrt 2 } {\frac{1}{2}.\left( {\frac{1}{{1 - t}} + \frac{1}{{1 + t}}} \right)dt} \\
= \frac{1}{2}.\ln \left| {\left( {1 - t} \right)\left( {1 + t} \right)} \right|_0^{1/\sqrt 2 }\\
= - \frac{1}{2}\ln 2
\end{array}$
