Find the area enclosed by a circle $r=4\sin\theta$ and out of $r^2=8\cos 2\theta$
I have tried the following integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\int_{\sqrt{8\cos2\theta}}^{4\sin\theta}dr d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int_{\sqrt{8\cos2\theta}}^{4\sin\theta}dr d\theta$ but it seems to be wrong
Consider full circle and individual parts separately, which are easy. There is no real need to lump them into a single integral in situations of area overlaps like here.
$$ r_{C}=4\sin\theta ; \; r_{ L}^2=8\cos 2\theta ; $$
$$ A_{Circle} =\frac12\int_{0}^{\frac{\pi}{6}} r_C^2 \cdot d\theta$$
$$A_{Lemniscate} =\frac12\int_{\pi/6}^{\pi/4} r_L^2 \cdot d\theta$$
$$ A_{Full Circle}= \pi 2^2 $$
$$ A_{Shaded\,Area} = A_{Full Circle} - 2 ( A_{Circle} + A_{Lemniscate} ) . $$