Q. $$\int \sin\left\{2\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)\right\}dx$$
$\implies$ $$\int \sin\left\{\sin ^{-1}\left(\frac{2\left(\sqrt{\frac{3-x}{3+x}}\right)}{1+\left(\sqrt{\frac{3-x}{3+x}}\right)^2}\right)\right\}dx$$
$\implies$ $$\int \:\frac{\left(\sqrt{3^2-x^2}\right)}{3}dx$$ $\implies$ANSWER $=$$$\frac{1}{3}\left(\left[\frac{x}{2}\sqrt{3^2-x^2}\right]+\left[\frac{9}{2}\sin ^{-1}\left(\frac{x}{3}\right)\right]\right)=\left(\left[\frac{x}{6}\sqrt{3^2-x^2}\right]+\left[\frac{3}{2}\sin ^{-1}\left(\frac{x}{3}\right)\right]\right)$$
Am I right?
Set $u=\arctan\sqrt{\dfrac{3-x}{3+x}}$. Then $$ \sqrt{\dfrac{3-x}{3+x}}=\tan u $$ and $$ \dfrac{3-x}{3+x}=\tan^2u $$ and, with straighforward computations, $$ x=3\cos2u $$ so $$ dx=-6\sin2u\,du $$ Thus your integral becomes $$ \int-6\sin^{2}2u\,du $$ that you should be able to carry over.