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I want to integrate

$$\int_{-1}^{1}\frac{dx}{x^3\sqrt{1-x^2}}.$$

I'm not sure where to begin as I have tried integrating by parts but end up in a continuous circle

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    $\begingroup$ You changed the limits of integration. It was from $-1$ to $1$. $\endgroup$ – Mhenni Benghorbal Nov 27 '13 at 16:33
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    $\begingroup$ Now you've changed the limits back. I note in my answer that this integral does not converge absolutely, but its Cauchy Principal Value is given in $(2)$ from my answer to be $0$. $\endgroup$ – robjohn Nov 27 '13 at 19:30
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Hint: Try the substitution $u=\sqrt{1-x^2}$.

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  • $\begingroup$ Having tried this i'm still not sure where to go! $\endgroup$ – Lauren Nov 27 '13 at 16:55
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    $\begingroup$ @Lauren: $$\begin{align} \int\frac{\mathrm{d}x}{x^3\sqrt{1-x^2}} &=\int\frac{\mathrm{d}\sqrt{1-u^2}}{\sqrt{1-u^2}^3u}\\ &=-\int\frac{\mathrm{d}u}{(1-u^2)^2} \end{align}$$ Now try partial fractions $\endgroup$ – robjohn Nov 27 '13 at 16:58
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    $\begingroup$ @robjohn okay ive understood to where you are, however when completing partial fractions and having A/(1-u^2)^2 and B/(1-u^2) B=0 which gives you the equation you began trying to split? $\endgroup$ – Lauren Nov 27 '13 at 18:31
  • $\begingroup$ @Lauren: to go any further I will need to add an answer; it's too long for a comment... $\endgroup$ – robjohn Nov 27 '13 at 18:34
  • $\begingroup$ You will want to use $1-u^2=(1-u)(1+u)$, I think. $\endgroup$ – Harald Hanche-Olsen Nov 27 '13 at 18:34
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The way I would approach this integral is by using trigonometric substitution

$$\int_{a}^{b}\frac{dx}{x^3\sqrt{1-x^2}}.$$ $$x=sin(θ)$$ $$dx=cos(θ)dθ$$ $$\int_{a}^{b}\frac{dθ}{sin^3(θ)\sqrt{1-sin^2(θ)}}.$$ you can simplify $$1-sin^2(θ)=cos^2(θ)$$ $$√(cos^2(θ))=cos(θ)$$ $$\int_{a}^{b}\frac{dθ}{sin^3(θ)\cos(θ)}.$$ Well this is the same as $$\int_{a}^{b}csc^3(θ)dθ.$$ The is a simple integration by parts. $$\int_{a}^{b}udv=uv|_{a}^{b}-\int_{a}^{b}vdu.$$ so we have to step up the substitutions again. $$u=csc(θ)$$ $$du=-csc(θ)cot(θ)$$ $$dv=csc^2(θ)$$ $$v=-cot(θ)$$ $$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(-cot(θ))(-csc(θ)cot(θ))dθ.$$ $$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc(θ)cot^2(θ))dθ.$$ we could rewrite $$cot^2(θ)=csc^2(θ)-1$$ because we know that $$1+cot^2(θ)=csc^2(θ)$$ $$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc(θ)(csc^2(θ)-1))dθ.$$ $$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}((csc^3(θ)-csc(θ))dθ.$$ we can just add the intergral of $$csc^3(θ)$$ to the left side and we get 2. $$\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}-\int_{a}^{b}(csc^3(θ))+\int_{a}^{b}csc(θ)dθ.$$ $$2\int_{a}^{b}csc^3(θ)dθ=-csc(θ)cot(θ)|_{a}^{b}+\int_{a}^{b}csc(θ)dθ.$$ so we can just the 2 from the left side. and Voilà. we get the $$\int_{a}^{b}csc^3(θ)dθ$$ $$\int_{a}^{b}csc^3(θ)dθ=\frac{1}{2}(-csc(θ)cot(θ))|_{a}^{b}+\frac{1}{2}\int_{a}^{b}csc(θ)dθ.$$. $$\frac{1}{2}\int_{a}^{b}csc(θ)dθ-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$. we make my integration a bit simpler i am going to make an other substitution. but before that i am going to multiply numerator and denominators of $$csc(θ)$$ by $$cot(θ)+csc(θ)$$. Which will give us. $$\frac{1}{2}\int_{a}^{b}-\frac{-csc^2(θ)-cot(θ)csc(θ)}{cot(θ)+csc(θ)}dθ-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$. Now i am going to substitute $$ z=cot(θ)+csc(θ) $$ and $$ dz=-csc^2(θ)-cot(θ)csc(θ)dθ $$ $$-\frac{1}{2}\int_{a}^{b}\frac{1}{z}dz-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$. this integral is going to result to -> $$-\frac{1}{2}\ln(z)|_{a}^{b}-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$. then we can back substitute z=$$cot(θ)+csc(θ)$$ which will give us. $$-\frac{1}{2}\ln(cot(θ)+csc(θ))|_{a}^{b}-\frac{1}{2}csc(θ)cot(θ))|_{a}^{b}$$. Then we can back substitute θ as well. $$θ=arcsin(x)$$ Then we get. $$-\frac{\sqrt{1-x^2}}{2x^2}|_{a}^{b}-\frac{1}{2}\ln\frac{\sqrt{1-x^2}+1}{x}|_{a}^{b} $$ we can simplify this to $$ -\frac{\sqrt{1-x^2}+x^2ln\frac{\sqrt{1-x^2}+1}{x}}{2x^2}|_{a}^{b} $$ and we can simplify this further more. $$ -\frac{\sqrt{1-x^2}+x^2(arcsech(x))}{2x^2}|_{a}^{b} $$

the final answer is. $$ \frac{1}{2}(\frac{-\sqrt{1-x^2})}{x^2}-ln(\sqrt{1-x^2}+1)+ln(x))|_{a}^{b} $$

just plug in b and a . minus it from each other. and thats the final answer

$$ \frac{1}{2}((\frac{-\sqrt{1-b^2})}{b^2}-ln(\sqrt{1-b^2}+1)+ln(b))-(\frac{-\sqrt{1-a^2})}{a^2}-ln(\sqrt{1-a^2}+1)+ln(a))) $$

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  • $\begingroup$ query at the start of this, when you have dx=cosθdθ, surely this should replace dx, but in your workings the cosθ isnt there, only the dθ? $\endgroup$ – Lauren Nov 27 '13 at 19:13
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    $\begingroup$ its because $$ \sqrt{1-sin^2(θ)} $$ is equal to $$ cos(θ) $$. because $$ sin^2(θ)+cos^2(θ)=1 $$. so when you mess around with the equation you get $$ 1-sin^2(θ) = cos^(θ) $$ . $$ \sqrt(1-sin^2(θ) = cos(θ) $$. I hope this helped if not comment i will reply asap . $\endgroup$ – KGTW Nov 27 '13 at 20:06
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    $\begingroup$ $$ x=sin(θ) $$ the derivative of this is equal to $$ dx=cos(θ)dθ $$. So we can substitute these variables. because in the equation that we're working on have these equivalent variables are there present for so we can proceed with the substitution process. $\endgroup$ – KGTW Nov 27 '13 at 20:09
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    $\begingroup$ I am sorry i made a typo mistake above. the correction is $$ 1-sin^2(θ)=cos^2(θ) $$ $${\sqrt{1-sin^2(θ)}=cos(θ)} $$ $\endgroup$ – KGTW Nov 27 '13 at 20:32
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    $\begingroup$ I see you have changed the limits of integretion. From a and b to 1 and -1. All you have to do it apply b=1 and a=-1. And do some algebraic simplification and you will get the solution. $\endgroup$ – KGTW Nov 27 '13 at 22:57
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Use $x=\sqrt{1-u^2}$ $$ \begin{align} \int\frac{\mathrm{d}x}{x^3\sqrt{1-x^2}} &=\int\frac{\mathrm{d}\sqrt{1-u^2}}{\sqrt{1-u^2}^3u}\\ &=-\int\frac{\mathrm{d}u}{(1-u^2)^2}\\ &=-\frac14\int\left(\frac1{(1-u)^2}+\frac1{(1+u)^2}+\frac1{1+u}+\frac1{1-u}\right)\,\mathrm{d}u\\ &=C-\frac14\left(\frac1{1-u}-\frac1{1+u}+\log(1+u)-\log(1-u)\right)\\ &=C-\frac14\left(\frac{2u}{1-u^2}+2\log\left(\frac{1+u}{\sqrt{1-u^2}}\right)\right)\\ &=C-\frac12\left(\frac{\sqrt{1-x^2}}{x^2}+\log\left(\frac{1+\sqrt{1-x^2}}{x}\right)\right)\tag{1} \end{align} $$ The integral above was done assuming that $x\gt0$. However, this can easily be extended by noticing that the integrand is an odd function, therefore, the integral is an even function. That is, the following is valid for all $x$: $$ \int\frac{\mathrm{d}x}{x^3\sqrt{1-x^2}} =C-\frac12\left(\frac{\sqrt{1-x^2}}{x^2}+\log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right)\right)\tag{2} $$ If your domain of integration spans the origin, the integral won't converge absolutely, but the Cauchy Principal Value will be given by $(2)$.


The question has been changed so that the domain of integration spans the origin. As mentioned above, the integral now does not converge absolutely. However, the Cauchy Principal Value is $0$.

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