Is this a trig substitution or a u-substitution problem?

It dont seem to work out for trig substitution..

$\displaystyle \int \frac {x^3}{ \sqrt {1 - 4x^2}}dx$

$\displaystyle x = sin\theta$

$\displaystyle dx = cos\theta d\theta$

$\displaystyle \sqrt{1-4x^2}$

$\displaystyle \sqrt{2^2(1 - sin^2\theta)}$

$\displaystyle = 2cos\theta$

sub back in

$\displaystyle \int \frac {sin^3\theta}{2 cos\theta}$

is this correct?