We have,
∫sin2x√9−sin4xdx
Now,
∫sin2x√32−(sin2x)2dx
Let
sin2x=t
ddxsin2x=dt
2sinxcosxdx=dt
sin2xdx=dt
∫dt√32−t2
Using that,
∫1√a2−x2dx=sin−1xa
∫dt√32−t2=sin−1t3+C
∫dt√32−t2=sin−1sin2x3+C
∫√sin4xdx