Cho I = intlimits_0fracpi a fracsin 2x1 + sin 2xdx , với giá trị nguyên nào của

$\begin{array}{l}
I = \int\limits_0^{\frac{\pi }{a}} {\frac{{\sin 2x}}{{1 + {{\sin }^2}x}}dx} \\
= \int\limits_0^{\frac{\pi }{a}} {\frac{{d\left( {1 + {{\sin }^2}x} \right)}}{{1 + {{\sin }^2}x}}} = \left. {\ln \left( {1 + {{\sin }^2}x} \right)} \right|_0^{\frac{\pi }{a}} = \ln \left( {1 + {{\sin }^2}\left( {\frac{\pi }{a}} \right)} \right) = \ln 2 \Leftrightarrow {\sin ^2}\left( {\frac{\pi }{a}} \right) = 1 \Leftrightarrow \left[ \begin{array}{l}
\sin \frac{\pi }{a} = 1\\
\sin \frac{\pi }{a} = - 1
\end{array} \right.
\end{array}$ $\frac{\pi }{a} = \pm \frac{\pi }{2} + k2\pi ,\,\,\,k \in \mathbb{Z} \Rightarrow a = \frac{2}{{4k \pm 1}}$. Vậy $a$ nguyên $ \Leftrightarrow k = 0 \Rightarrow a = \pm 2.$