Evaluating $\int\sqrt{\frac{2-x}{x-3}} dx$ 
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$

Need help in spotting my mistake: 
$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
$x-2 = t^2$
$\implies dx = 2t dt$
$$2 \int \sqrt{\dfrac{1}{1-t^2}} t^2dt$$
$t= \sin \theta $
$dt = \cos\theta d\theta$
$$\int (1- \cos 2\theta )d\theta $$
$=\theta - \dfrac{\sin 2\theta}{2}$
$$= \arcsin(t)- t\sqrt{1-t^2}$$
$= \arcsin (\sqrt{x-2})- \sqrt{(x-2)(3-x)}+C$
But answer given is: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$

My HINT: Substituting the square root we get
$$t=\sqrt{\frac{2-x}{x-3}}$$
then we get $$x=\frac{2+t2}{t^2+1}$$ and $$dx=-\frac{2t}{(t^2+1)^2}dt$$
Try this!
