$=\int_0^\pi \frac{(\pi-\theta) \sin (\pi-\theta)}{1+\cos ^2(\pi-\theta)} d \theta$

On adding Eqs. (i) and (ii), we get
$2 I=\int_0^\pi \frac{\pi \sin \theta}{1+\cos ^2 \theta} d \theta$
Let $\cos \theta=t \Rightarrow-\sin \theta d \theta=d t$
$\begin{aligned}
\therefore \quad 2 I & =-\pi \int_1^{-1} \frac{1}{1+t^2} d t \\
& =2 \pi \int_0^1 \frac{1}{1+t^2} d t \\
& =2 \pi\left[\tan ^{-1} t\right]_0^1=2 \pi \cdot \frac{\pi}{4}
\end{aligned}$
$\Rightarrow \quad I=\frac{\pi^2}{4}$