Solve: $\int\frac{\sin 2x}{\sqrt{3-(\cos x)^4}}$ (and a question about $t=\tan \frac{x}{2}$) I tried substituting $$t=\tan \frac{x}{2}$$ but the nominator is $\sin {2x}$, so is there a way to get from $$\sin x=\frac{2x}{1+x^2}$$ to an expression with $\sin 2x$?

HINT:
As $\int\sin2x\ dx=-\dfrac{\cos2x}2=\dfrac{1-2\cos^2x}2$
Set $\cos^2x=u$
$$\int\frac{\sin(2x)}{\sqrt{3-\cos^4(x)}}\ dx=-\int\dfrac{du}{\sqrt{3-u^2}}$$
Now for $a>0,$ $\dfrac{d(\arcsin x/a)}{dx}=\dfrac1{\sqrt{a^2-x^2}}$
