Find `int(sin2xcos2xdx)/(sqrt(9-cos^4(2x)))`

Explanation

`int( sin2xcos2xdx)/sqrt(9-cos^4(2x)}dx `
Let `cos^2(2x)=t`
Differentiating both sides w.r.t.x
`-2.2 cos2x sin2x =dt/dx`
`dx=dt/(-4cos2xsin2x)`
Hence our equation becomes ,
`int (sin2xcos2x)/(sqrt(9-cos^4(2x)))dx=int (sin2xcos2x)/sqrt(9-t^2) dx`
`=int(sin2xcos2x)/sqrt(9-t^2) times dt/(-4cos2x sin2x)`
`=-1/4 int dt/{sqrt[(3)^2-(t)^2]}`
`=-1/4 [sin^(-1)(t/3)+C_1]`
`=-1/4sin^(-1) (t/3)-C_1/4`
`=-1/4sin^-1[1/3cos^2(2x)]+C`