Lời giải

Đặt \( t = \pi - x \quad \Rightarrow \quad dt = -dx\)

\(\begin{cases}
x = 0 \Rightarrow t = \pi \\
x = \pi \Rightarrow t = 0
\end{cases}\)

\(I = \int_\pi^0 \frac{(\pi - t) \sin (\pi - t)}{1 + \cos^2 (\pi - t)} (-dt)\)

   \(= \pi \int_0^\pi \frac{ \sin t}{1 + \cos^2 t} dt - \int_0^\pi \frac{t \sin t}{1 + \cos^2 t} dt\)

\(\Rightarrow I = \frac{\pi}{2} \int_0^\pi \frac{\sin t}{1 + \cos^2 t} dt \quad\)    Đặt  \(u = \cos t\)

         \(= \frac{\pi}{2} \int_{-1}^1 \frac{1}{1 + u^2} \, du \quad \text{Đặt } u = \tan v \quad -\frac{\pi}{2} < v < \frac{\pi}{2}\)

          \(= \frac{\pi^2}{4}\)