Busy Integral

$\text{Taking}\space \sin^2x \space \text{common from the denominator}$ $\text{and using}\color{#3D99F6}{\dfrac{1}{\sin^2x}=\csc^2x, csc^2x=1+cot^2x}$ $I= \displaystyle \int_{0}^{\pi}\dfrac{dx}{5+3\cos^2x}=\displaystyle \int_{0}^{\pi}\dfrac{(\csc^2 x) \text{dx}}{5\csc^2x+3\cot^2x}$ $=\displaystyle \int_{0}^{\pi}\dfrac{(\csc^2x) \text{dx}}{5+8cot^2x}$ Notice d(cotx)= $-csc^2x$ and using $\color{#3D99F6}{\displaystyle \int \frac{\text{dx}}{x^2+a^2}=\frac{1}{a}tan^{-1}(\frac{x}{a})}$ $I=\dfrac{-1}{2\sqrt{10}}[ tan^{-1} (\frac{2\sqrt{2}cotx}{\sqrt{5}})]^\pi_0$ $=\color{magenta}{\frac{-1}{2\sqrt{10}}(-\frac{\pi}{2}-\frac{\pi}{2})}$ $=\color{limegreen}{\frac{\pi}{2\sqrt{10}}}$ a=2,b=10 and a+b= $\Large 12$