<s>[INST] 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: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$
 [/INST] $$\sqrt{\dfrac{2-x}{x-3}}=\sqrt{\dfrac{x-2}{3-x}}$$
For real calculus, $2\le x<3$
$\iff2-\dfrac{2+3}2\le x-\dfrac52<3-\dfrac52$
Choose $x-\dfrac52=\dfrac{\cos2t}2$
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