$$\int\sqrt{\dfrac{2-x}{x-3}} dx$$
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: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$
Find the derivatives of both answers. You will find that yours is the correct one.
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.
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!
$$\sqrt{\dfrac{2-x}{x-3}}=\sqrt{\dfrac{x-2}{3-x}}$$
For real calculus, $2\le x<3$
$\iff2-\dfrac{2+3}2\le x-\dfrac52<3-\dfrac52$
Choose $x-\dfrac52=\dfrac{\cos2t}2$
