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 $\sin 2x$?
 A: $$\int\frac{\sin(2x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$

Use $\sin(2x)=2\sin(x)\cos(x)$:

$$2\int\frac{\sin(x)\cos(x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$

Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:

$$-2\int\frac{u}{\sqrt{3-u^4}}\space\text{d}u=$$

Substitute $s=u^2$ and $\text{d}s=2u\space\text{d}u$:

$$-\int\frac{1}{\sqrt{3-s^2}}\space\text{d}s=-\int\frac{1}{\sqrt{3}\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=-\frac{1}{\sqrt{3}}\int\frac{1}{\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=$$

Substitute $p=\frac{s}{\sqrt{3}}$ and $\text{d}p=\frac{1}{\sqrt{3}}\space\text{d}s$:

$$-\int\frac{1}{\sqrt{1-p^2}}\space\text{d}p=-\arcsin\left(p\right)+\text{C}=-\arcsin\left(\frac{s}{\sqrt{3}}\right)+\text{C}=$$
$$-\arcsin\left(\frac{u^2}{\sqrt{3}}\right)+\text{C}=-\arcsin\left(\frac{\cos^2(x)}{\sqrt{3}}\right)+\text{C}$$
A: 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 $a>0,$ $\dfrac{d(\arcsin x/a)}{dx}=\dfrac1{\sqrt{a^2-x^2}}$
