12
$\begingroup$

I am trying to evaluate the indefinite integral of

$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$

The first thing I did was the substitution rule: $u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to $$\int \sqrt{\frac{2-u}{u}} \, \frac{du}2$$ or $$\frac 12 \int \sqrt{\frac 2u - 1} \, du$$ I'm a bit stuck here. May I ask for help on how to proceed?

$\endgroup$
2
  • $\begingroup$ Have you tried $u=1-x^2$ ? $\endgroup$
    – Shabbeh
    Aug 5, 2014 at 5:23
  • $\begingroup$ If you search it on Wolfram, you end up needing $\arcsin(x)$. I've tried starting the substitution with $u=\arcsin(x)$ but it gets messy quite quickly. $\endgroup$
    – Jam
    Aug 5, 2014 at 5:37

5 Answers 5

8
$\begingroup$

You had reached $$\int \frac{1}{2}\sqrt{\frac{2-u}{u}}\,du.$$ Now a substitution that works well is $u=2\sin^2 t$.

$\endgroup$
3
  • $\begingroup$ Substituting $u=\sin^2 t$ and $du = 4 \sin t \cos t \, dt$, I reduced my integral to $$2 \int \cos^2 t \, dt=\int 1 + \cos (2t) \, dt.$$ Is this correct so far by any chance? If so, I think I can finish it off from here, although the antiderivative with all the $\arcsin$'s can get ugly. $\endgroup$
    – Cookie
    Aug 5, 2014 at 6:09
  • $\begingroup$ ^^Typo in the first line of my above comment: I meant to say $u = 2 \sin^2 t$, which is what you said in your answer. $\endgroup$
    – Cookie
    Aug 5, 2014 at 6:15
  • $\begingroup$ Yes, apart from constants that I am not good at, we do need to integrate $\cos^2 t$. A standard way is the double angle formula, though I like integration by parts. $\endgroup$ Aug 5, 2014 at 6:16
7
$\begingroup$

Use the substitution $x^2=\cos(2\theta) \Rightarrow x\,dx=-\sin(2\theta)\,d\theta$ i.e $$I=\int x\sqrt{\frac{1-x^2}{1+x^2}}\,dx=-\int \sin(2\theta)\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} \, d\theta=-\int \sin(2\theta)\tan\theta\,d\theta$$ where I used $\cos(2\theta)=2\cos^2\theta-1=1-2\sin^2\theta$. $$\Rightarrow I=-2\int \sin^2\theta\,d\theta$$ The last integral is easily handled by $\sin^2\theta=(1-\cos(2\theta))/2$.

$\endgroup$
4
$\begingroup$

$\bf{My\; Solution::}$ Let $\displaystyle I = \int x\sqrt{\frac{1-x^2}{1+x^2}}dx\;,$ Now Substitute $x^2=t\;,$ Then $\displaystyle xdx = \frac{dt}{2}$

So $\displaystyle I = \frac{1}{2}\int \sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int \frac{\left(1-t\right)}{\sqrt{1-t^2}}dt =\frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{1}{2}\int \frac{t}{\sqrt{1-t^2}}dt$

So $\displaystyle I = \frac{1}{2}\sin^{-1}(t)-\frac{1}{2}\sqrt{1-t^2}+\mathcal{C} = \frac{1}{2}\sin^{-1}(\sqrt{x})-\frac{1}{2}\sqrt{1-x}+\mathcal{C}$

$\endgroup$
4
$\begingroup$

Another way to solve the integral is as follows. Let $u=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$ so that

$$u^{2}(1+x^{2})=1-x^{2}$$

$$u^{2}-1=-x^{2}(1+u^{2})$$

$$x^{2}=\frac{1-u^{2}}{1+u^{2}}$$

so

$$2xdx=\frac{(-2u)(1+u^{2})-(2u)(1-u^{2}))}{(1+u^{2})^{2}}du=\frac{-4u}{(1+u^{2})^{2}}du$$

Going back to the integral we have:

$$\int x\sqrt{\frac{1-x^{2}}{1+x^{2}}}dx=\frac{1}{2}\int\sqrt{\frac{1-x^{2}}{1+x^{2}}}2xdx=\int u\frac{-2u}{(1+u^{2})^{2}}du=\frac{u}{1+u^{2}}-\int\frac{1}{1+u^{2}}du$$

$\endgroup$
0
$\begingroup$

$\text {Let } x^{2}=2 \sin ^{2} \theta-1 \textrm{ for } \frac{\pi}{4} \leqslant \theta \leqslant \frac{\pi}{2}, \text {then } x d x=\sin \theta \cos \theta d \theta $.

\begin{aligned}\int x \sqrt{\frac{1-x^{2}}{1+x^{2}}} d x&=\int \sqrt{\frac{2-2 \sin ^{2} \theta}{2 \sin ^{2} \theta} }\sin \theta \cos \theta d \theta \\ &=\int \cos ^{2} \theta d \theta\\&=\frac{1}{2} \int(1+\cos 2 \theta) d \theta \\ &=\frac{1}{2} \theta+\frac{\sin 2 \theta}{4}+C\\&=\frac{1}{2} \sin ^{-1} \sqrt{\frac{x^{2}+1}{2}}+\frac{1}{4} \sqrt{1-x^{4}}+C \end{aligned}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .