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

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: $-\arcsin(2x-5)+ \sqrt{(2-x)(x-3)}$

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    $\begingroup$ Should be: $2-x =-t^2$ For starters. Check your signs. $\endgroup$ – JavaMan Aug 3 '18 at 13:13
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    $\begingroup$ @JavaMan Please see carefully there's no problem there. $\endgroup$ – Archer Aug 3 '18 at 13:16
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Find the derivatives of both answers. You will find that yours is the correct one.

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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.

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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!

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$$\sqrt{\dfrac{2-x}{x-3}}=\sqrt{\dfrac{x-2}{3-x}}$$

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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