<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] Your answer is absolutely fine, as you can check by differentiating. 
If the given answer were correct as well, the equations
\begin{align}
C &= \frac{\pi}{2} \, , \\
\frac{\pi}{2} + C &= - \frac{\pi}{2} \, ,
\end{align}
which we obtain by evaluating both answers at $x=2$ and $x=3$, would have to be satisfied simultaneously for some $C \in \mathbb{R}$ . Clearly, this is impossible, so the given answer must be wrong.
</s>