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Đặt: $t = \pi - x \Rightarrow dt = - dx$
Đổi cận: $\begin{cases} x=0 \rightarrow t=\pi \\ x=\pi \rightarrow t=0 \end{cases}$
$\begin{array}{l} \Rightarrow I =- \int\limits_\pi^0 {\frac{{\left( {\pi - t} \right)\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt= \int\limits_0^\pi {\frac{{\left( {\pi - t} \right)\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt = \pi \int\limits_0^\pi {\frac{{\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt - I\\
\Rightarrow 2I = \pi \int\limits_0^\pi {\frac{{\sin t}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} dt = - \pi \int\limits_0^\pi {\frac{{d(\cos t)}}{{1 + c{\rm{o}}{{\rm{s}}^2}t}}} = -\pi \arctan \cos t\left|\begin{array}{I}\pi\\0\end{array}\right.=\pi \left( {\frac{\pi }{4} +\frac{\pi }{4}} \right)\\
\Rightarrow I =\frac{\pi^2}{4}.
\end{array}$
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