Hello,
If $0<a<b$, then we have $\frac12\displaystyle\int_0^{2\pi} \frac{d\theta}{(b+a*cos(\theta))^2}=\frac{\pi b}{(b^2-a^2)^{\frac32}}\tag{1}$

Now, the LHS of this equation is the area of ellipse written as $\frac12 \displaystyle\int_0^{2\pi}\rho(\theta)^2\, d\theta$

I want to know 'HOW'?

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