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I need to solve:

$$\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$

What I did is:

Substitute: $x=2\cos^2 \theta + 3\sin^2 \theta$. Now:

$$\begin{align} x &= 2 - 2\sin^2 \theta + 3 \sin^2 \theta \\ x &= 2+ \sin^2 \theta \\ \sin \theta &= \sqrt{x-2} \\ \theta &=\sin^{-1}\sqrt{x-2} \end{align}$$

and, $ cos \theta = \sqrt{(3-x)} $

$ \theta=\cos^{-1}\sqrt{(3-x)}$

The integral becomes:

$$\begin{align} &= \int{\sqrt[]{\frac{2\cos^2 \theta + 3\sin^2 \theta-3}{2-2\cos^2 \theta - 3\sin^2 \theta}} ~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta}\\ % &= \int{\sqrt[]{\frac{2\cos^2 \theta + 3(\sin^2 \theta-1)}{2(1-\cos^2 \theta) - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta)}~{\rm d}{\theta} \\ % &= \int\sqrt[]{\frac{2\cos^2 \theta - 3\cos^2 \theta}{2\sin^2 \theta - 3\sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\ % &= \int\sqrt[]{\frac{-\cos^2 \theta }{- \sin^2 \theta}}~~(2 \cos \theta\sin\theta) ~{\rm d}\theta \\ % &= \int \frac{\cos \theta}{\sin\theta}~~(2 \cos \theta\sin\theta)~{\rm d}\theta \\ % &= \int 2\cos^2 \theta~{\rm d}\theta \\ % &= \int (1- \sin 2\theta)~{\rm d}\theta \\ % &= \theta - \frac {\cos 2\theta}{2} + c \\ % &= \sin^{-1}\sqrt{x-2} - \frac {\cos 2(\sin^{-1}\sqrt{x-2})}{2} + c \end{align}$$

But, The right answer is :

$$\sqrt{\frac{3-x}{x-2}} - \sin^{-1}\sqrt{3-x} + c $$

Where am I doing it wrong?

How do I get it to the correct answer??

UPDATE:

I am so sorry I wrote:

= $\int 2\cos^2 \theta .d\theta$

= $\int (1- \sin 2\theta) .d\theta$

It should be:

= $\int 2\cos^2 \theta .d\theta$

= $\int (1+ \cos2\theta) .d\theta$

= $ \theta + \frac{\sin 2\theta}{2} +c$

What do I do next??

UPDATE 2:

= $ \theta + \sin \theta \cos\theta +c$

= $ \theta + \sin \sin^{-1}\sqrt{(x-2)}. \cos\cos^{-1}\sqrt{(3-x)}+c$

= $ \sin^{-1}\sqrt{(x-2)}+ \sqrt{(x-2)}.\sqrt{(3-x)}+c$

Is this the right answer or I have done something wrong?

user63762453
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  • Ali.E. told you what was the mistake. In any manner, if I may suggest, when you have computed an antiderivative, compute its derivative to check if you get back to the integrand. Use WA (only foir that !). Cheers. – Claude Leibovici Jan 03 '15 at 09:14
  • By the way, you may suppose $x= 2+ \sin^2 \theta$ from the beginning. $x=2\cos^2 \theta + 3\sin^2 \theta$ isn't necessary! –  Jan 03 '15 at 09:32
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    I get the result $\sqrt{(x-2)(3-x)}+\arcsin(\sqrt{x-2})+C$. And, differentiating what you say is the right answer don't get me to $\sqrt{(x-3)/(2-x)}$. By the way, for what values of $x$ do you expect this to be real? – mickep Jan 03 '15 at 09:45
  • After your update: What you do next is to use the formula $\sin 2z=2\sin z\cos z$. Then simplify. – mickep Jan 03 '15 at 09:47
  • I will do that and get back here. Thanks :) – user63762453 Jan 03 '15 at 09:49
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    @ mickep: It does not matter if it is real, you can simply factor $i=\sqrt(-1)$ to make the rest real! Differentiation of $\sqrt{(3-x)(x-2)} - \sin^{-1} \sqrt{3-x} + c$ results in $\sqrt{(x-3)/(2-x)}$. –  Jan 03 '15 at 10:03
  • You have now the right answer, yes. @ali-e I agree, but at the point one struggles with these kind of calculations, one is usually doing real analysis. I was perhaps wrong though, to assume that was the case here (there is no such tag)... – mickep Jan 03 '15 at 10:10
  • @mickep I think the correct answer I have mentioned in the question is wrong. It was in a book. – user63762453 Jan 03 '15 at 10:14
  • Indeed, it was wrong. – mickep Jan 03 '15 at 10:15
  • https://math.stackexchange.com/questions/4816777/evaluate-the-integral-int-sqrt-frac2x13x2dx – Integreek Nov 22 '24 at 14:45

6 Answers6

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Hint:

For evaluating the integral, notice that $$t^2 = \frac{x-3}{2-x} \implies \mathrm dx = -\frac{2t}{(t^2+1)^2}\mathrm dt.$$ Hence,

$$\int \sqrt{\frac{x-3}{2-x}}\mathrm dx = -2\int\frac{t^2}{(t^2+1)^2}\mathrm dt$$

Integreek
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Alex Silva
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  • I didn't get you. – user63762453 Jan 03 '15 at 09:46
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    @Nivedita: He's telling you a simpler route than what you've taken $$\text{Let } t^2 = \frac{x - 3}{2-x} = \frac{x-2 -1}{-(x-2)} = \frac{1}{x-2} - 1$$ Do simple substitution and integrate. – Nick Jan 03 '15 at 09:52
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$$I=\int\sqrt{\frac{x-3}{2-x}}~{\rm d}x$$

Integrating Let $x=2\cos^2t+3\sin^2t$, $dx=\sin2tdt$

$$I=\int\sqrt{\frac{-\cos^2t}{-\sin^2t}}\sin2tdt=\int2\cos^2tdt=\int(1+\cos2t)dt=t+\frac12\sin2t+c\\I=\underbrace{\cos^{-1}\sqrt{3-x}}_{\pi/2-\sin^{-1}\sqrt{3-x}}+\sqrt{x-2}\sqrt{3-x}+c\\I=\underbrace{\sqrt{x-2}\sqrt{3-x}}_{\sqrt{5x-x^2-6}}-\sin^{-1}{\sqrt{3-x}}+c'$$

Differentiating back $$I'=\frac1{2\sqrt{(x-2)(3-x)}}\cdot(5-2x)-\underbrace{\frac1{\sqrt{1-(\sqrt{3-x})^2}}}_{\sqrt{x-2}}\cdot\frac1{2\sqrt{3-x}}(-1)\\I'=\frac{2(3-x)}{2\sqrt{(x-2)(3-x)}}=\sqrt{\frac{3-x}{x-2}}=\sqrt{\frac{x-3}{2-x}}$$

RE60K
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2

$2(\cos x)^2=1+\cos(2x)$ but $2(\cos x)^2\neq1-\sin(2x)$

Vincenzo Tibullo
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2

to shorten the working, since the function is only well-defined for $x \in (2,3)$ we may substitute $$ 3-x \to s $$ giving $$ I=\int\sqrt{\frac{x-3}{2-x}} .dx = - \int\sqrt{\frac{s}{1-s}} .ds $$ now substitute $$ s \to \sin^2 \theta \\ ds \to 2\sin\theta \cos\theta\cdot d\theta $$ so $$ I = -\int2\sin^2 \theta \cdot d\theta = \int (\cos 2\theta-1)\cdot d\theta=\frac12\sin2\theta-\theta+c \\ = \sqrt{s(1-s)} - \sin^{-1}s \\ =\sqrt{(3-x)(x-2)} - \sin^{-1} \sqrt{3-x} + c $$

David Holden
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Another approach is to perform the substitution $t=\sqrt{x-2}$ since its derivative is present in the integrand:

$$\begin{align}\int\sqrt{\frac{3-x}{x-2}}&=2\int\sqrt{1-t^2}\mathrm dt\end{align}$$

Now, just use the result of the standard integral

$$\int\sqrt{a^2-x^2}\mathrm dx=\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C$$

and back-substitute.

Integreek
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0

An alternative method is to rationalize the numerator:

$$\begin{align}\int\sqrt{\frac{3-x}{x-2}}\mathrm dx&=\int\frac{3-x}{\sqrt{-x^2+5x-6}}\mathrm dx\\&=\int\frac{\frac12(-2x+5)+\frac12}{\sqrt{-x^2+5x-6}}\mathrm dx\end{align}$$

It is evident how to proceed forward.

Integreek
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