6
$\begingroup$

$$\int \sqrt{\frac{x}{2-x}}dx$$

can be written as:

$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$

there is a formula that says that if we have the integral of the following type:

$$\int x^m(a+bx^n)^p dx,$$

then:

  • If $p \in \mathbb{Z}$ we simply use binomial expansion, otherwise:
  • If $\frac{m+1}{n} \in \mathbb{Z}$ we use substitution $(a+bx^n)^p=t^s$ where $s$ is denominator of $p$;
  • Finally, if $\frac{m+1}{n}+p \in \mathbb{Z}$ then we use substitution $(a+bx^{-n})^p=t^s$ where $s$ is denominator of $p$.

If we look at this example:

$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx,$$

we can see that $m=\frac{1}{2}$, $n=1$, and $p=\frac{-1}{2}$ which means that we have to use third substitution since $\frac{m+1}{n}+p = \frac{3}{2}-\frac{1}{2}=1$ but when I use that substitution I get even more complicated integral with square root. But, when I tried second substitution I have this:

$$2-x=t^2 \Rightarrow 2-t^2=x \Rightarrow dx=-2tdt,$$

so when I implement this substitution I have:

$$\int \sqrt{2-t^2}\frac{1}{t}(-2tdt)=-2\int \sqrt{2-t^2}dt.$$

This means that we should do substitution once more, this time:

$$t=\sqrt{2}\sin y \Rightarrow y=\arcsin\frac{t}{\sqrt{2}} \Rightarrow dt=\sqrt{2}\cos ydy.$$

So now we have:

\begin{align*} -2\int \sqrt{2-2\sin^2y}\sqrt{2}\cos ydy={}&-4\int\cos^2ydy = -4\int \frac{1+\cos2y}{2}dy={} \\ {}={}& -2\int dy -2\int \cos2ydy = -2y -\sin2y. \end{align*}

Now, we have to return to variable $x$:

\begin{align*} -2\arcsin\frac{t}{2} -2\sin y\cos y ={}& -2\arcsin\frac{t}{2} -2\frac{t}{\sqrt{2}}\sqrt\frac{2-t^2}{2}={} \\ {}={}& -2\arcsin\frac{t}{2} -\sqrt{t^2(2-t^2)}. \end{align*}

Now to $x$:

$$-2\arcsin\sqrt{\frac{2-x}{2}} - \sqrt{2x-x^2},$$

which would be just fine if I haven't checked the solution to this in workbook where the right answer is:

$$2\arcsin\sqrt\frac{x}{2} - \sqrt{2x-x^2},$$

and when I found the derivative of this, it turns out that the solution in workbook is correct, so I made a mistake and I don't know where, so I would appreciate some help, and I have a question, why the second substitution works better in this example despite the theorem i mentioned above which says that I should use third substitution for this example?

$\endgroup$
  • 3
    $\begingroup$ Just a different thought, try u sub $\sqrt{2-x}=t$ and see what happens $\endgroup$ – imranfat Mar 8 '16 at 18:26
  • $\begingroup$ 2-x=t is better $\endgroup$ – Takahiro Waki Mar 8 '16 at 18:45
  • $\begingroup$ @imranfat: Isn't that just what he did? cdummie you are missing some square roots from the denominators of the arcsine arguments when going from $y$ to $t$, I guess, since $y=\arcsin\frac{t}{\sqrt2}$. Or, the roots in the last line are not supposed to be in the root. $\endgroup$ – MickG Mar 8 '16 at 20:47
  • $\begingroup$ @MickG Yeah, but I just tried to avoid the mumble jumble of that standard formula with that m,b,n,p . I became dizzy and so I had to close the page. I took a paper and did the u-sub I suggested and midway through, the integral became standard bread and butter, I quit... $\endgroup$ – imranfat Mar 8 '16 at 21:51
  • $\begingroup$ Seems the OP did his job correctly @imranfat: see my answer below. $\endgroup$ – MickG Mar 8 '16 at 21:53
3
$\begingroup$

Let me try do derive that antiderivative. You computed:

$$f(x)=\underbrace{-2\arcsin\sqrt{\frac{2-x}{2}}}_{f_1(x)}\underbrace{-\sqrt{2x-x^2}}_{f_2(x)}.$$

The easiest term is clearly $f_2$:

$$f_2'(x)=-\frac{1}{2\sqrt{2x-x^2}}\frac{d}{dx}(2x-x^2)=\frac{x-1}{\sqrt{2x-x^2}}.$$

Now the messier term. Recall that $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$. So:

\begin{align*} f_1'(x)={}&-2\frac{1}{\sqrt{1-\left(\sqrt{\frac{2-x}{2}}\right)^2}}\frac{d}{dx}\sqrt{\frac{2-x}{2}}=-2\frac{1}{\sqrt{1-\frac{2-x}{2}}}\cdot\frac{1}{\sqrt2}\frac{d}{dx}\sqrt{2-x}={} \\ {}={}&-2\sqrt{\frac2x}\cdot\frac{1}{\sqrt2}\cdot\frac{1}{2\sqrt{2-x}}\cdot(-1)=\frac{2}{\sqrt x}\frac{1}{2\sqrt{2-x}}=\frac{1}{\sqrt{2x-x^2}}. \end{align*}

So:

$$f'(x)=f_1'(x)+f_2'(x)=\frac{x}{\sqrt{2x-x^2}}=\frac{x}{\sqrt x}\frac{1}{\sqrt{2-x}}=\frac{\sqrt x}{\sqrt{2-x}},$$

which is your integrand. So you were correct after all! Or at least got the correct result, but no matter how I try, I cannot find an error in your calculations.

As for the book's solution, take your $f$, and compose it with $g(x)=2-x$. You get the book's solution, right? Except for a sign. But then $g'(x)=-1$, so the book's solution is also correct: just a different change of variables, probably, though I cannot really guess which.

$\endgroup$
  • $\begingroup$ Yeah, that's pretty much it... $\endgroup$ – imranfat Mar 8 '16 at 21:55
3
$\begingroup$

$$\int \sqrt{\frac{x}{2-x}}dx$$

Set $t=\frac {x} {2-x}$ and $dt=\left(\frac{x}{(2-x)^2}+\frac{1}{2-x}\right)dx$

$$=2\int\frac{\sqrt t}{(t+1)^2}dt$$

Set $\nu=\sqrt t$ and $d\nu=\frac{dt}{2\sqrt t}$

$$=4\int\frac{\nu^2}{(\nu^2+1)^2}d\nu\overset{\text{ partial fractions}}{=}4\int\frac{d\nu}{\nu^2+1}-4\int\frac{d\nu}{(\nu^1+1)^2+\mathcal C}$$

$$=4\arctan \nu-4\int\frac{d\nu}{(\nu^2+1)^2}$$

Set $\nu=\tan p$ and $d\nu=\sec^2 p dp.$ Then $(\nu^2+1)^2=(\tan^2 p+1)^2=\sec^4 p$ and $p=\arctan \nu$

$$=4\arctan \nu-4\int \cos^2 p dp$$

$$=4\arctan \nu-2\int \cos(2p)dp-2\int 1dp$$

$$=4\arctan \nu-\sin(2p)-2p+\mathcal C$$

Set back $p$ and $\nu$:

$$=\color{red}{\sqrt{-\frac{x}{x-2}}(x-2)+2\arctan\left(\sqrt{-\frac{x}{x-2}}\right)+\mathcal C}$$

$\endgroup$
  • $\begingroup$ You're welcome. I would advise you to add some steps to the last passage since it is not immediately clear to me how you reverted to the variable $x$. $\endgroup$ – MickG Mar 8 '16 at 21:10
  • $\begingroup$ Noting that $4\arctan\nu-2p=2\arctan\nu$ since $p=\arctan\nu$ and applying the formula $\sin(2\arctan(p))=\frac{2p}{p^2+1}$ to get this as an intermediate step for the $\sin(2p)$ term would be of great help to those (like me) trying to "set back $p$ and $\nu$" mentally. :) $\endgroup$ – MickG Mar 8 '16 at 21:30
1
$\begingroup$

Alternative solution - let $x=2t^2$, then

$$I=\int\sqrt{\frac{x}{2-x}}\mathrm{d}x=4\int\frac{t^2}{\sqrt{1-t^2}}\mathrm{d}t=4J$$

By parts we have

$$J=-t\sqrt{1-t^2}+\int\sqrt{1-t^2}\;\mathrm{d}t = -t\sqrt{1-t^2}+\int\frac{1-t^2}{\sqrt{1-t^2}}\;\mathrm{d}t\!=\!-t\sqrt{1-t^2}+\arcsin t-J $$

Hence

$$I=4J=2\cdot 2J =2\arcsin t -2t\sqrt{1-t^2} = 2\arcsin\sqrt{\frac{x}{2}}-\sqrt{2x-x^2} + C$$

The solutions are equivallent because of formula : $$\arcsin x= \frac{\pi}{2}-\arcsin{\sqrt{1-x^2}} $$

Clearly, take $\sin$ of both sides, with the fact that $\sin (\frac{\pi}{2}-x)=\cos x$ :

$$ x= \cos\arcsin{\sqrt{1-x^2}}=\sqrt{1-\sin^2{\arcsin{\sqrt{1-x^2}}}} =\sqrt{1-(1-x^2)} = x $$

$\endgroup$
0
$\begingroup$

Let $u=\sqrt{2-x}$ then we simply want

$-2\int \sqrt{2-u^2}du$ which is simple after $u=\sqrt{2}\sin{v}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.