Area above asymptote of polar curve

We can do a parametric integration of this curve shifted up by $1$

$x(\theta)=\dfrac{Cos(\theta)}{\theta-\pi}$

$y(\theta)=\dfrac{Sin(\theta)}{\theta-\pi}+1$

$\displaystyle \int _{ a }^{ b }{ y'(\theta )x(\theta )d\theta }= \dfrac { 2(\pi - \theta)+Sin(2\theta ) }{ { 4\left( \pi - \theta \right) }^{ 2 } }$

For $\theta = \dfrac{\pi}{2}$ to $\infty$ , this works out to $\dfrac{1}{\pi}$ , so the answer is twice that.