I want to calculate moment of inertia of uniform ring about an axis which is ${\large perpendicular}$ to the surface of ring and passing through $F$ using integration.
By parallel axis theorem, Moment of inertia of ring about this axis is $2MR^2$.
Let radius and mass of Ring be $R$ and $m$ respectively
Let $DE$ be the differential element of length $dx$ and mass $dm$. Let $\angle AFE$ be $\theta$ and $\angle EFD$ be $d\theta$. Also Let $EF=FD=x$
$$I=\int x^2 dm$$ , where $x$ represents perpendicular distance from axis of rotation.
As $$dm=\frac{m}{2\pi R}dx$$
$$\implies I=\frac{m}{2\pi R}\int x^2 dx$$
As $AF=AE=R$, From cosine law we have $cos\theta=\frac{x}{2R}$. Also $dx=x d\theta$.
$$I=\frac{m}{2\pi R}\int_{0}^{2\pi} 4R^2cos^2\theta(2Rcos\theta) d\theta$$
$$\ \implies I=\frac{4mR^2}{\pi} \int_{0}^{2\pi}cos^3 \theta d\theta$$
This gives $I=0$ but this is wrong as answer is $2mR^2$. I wasn't able to find mistake in calculation for moment of inertia. Initially I thought that problem is with limits of integrals but then I realise that even if limits are wrong that $\pi$ in denominator won't cancel which still makes our result wrong.
Note:- This question is now solved. The mistake lies in limits of integral. Angle should have been measured from centre of Ring.
Thank you for your help!