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I was asked to find the following integral:

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

What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$ And:$$x=\sqrt{\frac{1}{1-u^2}}$$ And we can replace: $$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\int\frac{du}{\sqrt{1-u^2}}=\arctan(u)+C=\arctan(\sqrt{1-\frac{1}{x^2}})+C$$ The problem is, when that result is derived we don't get the original expression. I just can't find my mistake, so some help would be appreciated.

Renascence_5.
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Jonathan Martinez
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    Hint: $x\rightarrow 1/x$ – tired Jan 14 '17 at 15:25
  • Thanks, that indeed solved the integral. But my main interest was to find the mistake in my solution, since I can't find it and I will surely make the same mistake later... – Jonathan Martinez Jan 14 '17 at 15:34
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    @tired Were you too "tired" to post this as a solution? (+1) for the comment. – Mark Viola Jan 14 '17 at 15:34
  • Your solution should be $\text{arscin}(u)$ not $\arctan(u)$. – Mark Viola Jan 14 '17 at 15:42
  • @Dr. MV in most cases i only post a solution if the problem poses some challenge to me, which wasn't the case here. Furhtermore the hint is good enough for finishing off this one quickly by the op himself – tired Jan 14 '17 at 15:46
  • @tired I was trying and failing to make a play of words from your user name. – Mark Viola Jan 14 '17 at 15:50
  • @Dr. MV i got the irony, no worries ;) Nevertheless i thought it is a good opportunity to explain my behaviour of commenting often and answering seldom – tired Jan 14 '17 at 16:12

3 Answers3

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Substitute $\text{u}=\frac{1}{x}$:

$$\mathcal{I}\left(x\right)=\int\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\space\text{d}x=-\int\frac{1}{\sqrt{1-\text{u}^2}}\space\text{d}\text{u}$$

Now, use:

$$\frac{\text{d}\arcsin\left(x\right)}{\text{d}x}=\frac{1}{\sqrt{1-x^2}}$$

So, we get:

$$\mathcal{I}\left(x\right)=\text{C}-\arcsin\left(\frac{1}{x}\right)=\text{C}-\text{arccsc}\left(x\right)$$

Jan Eerland
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The substitution $u=\sqrt{1-\frac{1}{x^2}}$ is equivalent to $|x|=\frac{1}{\sqrt{1-u^2}}$. For $x>1$, we have

$$\begin{align} \int \frac{1}{x^2\sqrt{1-\frac1{x^2}}}\,dx&=\int \frac{1}{\sqrt{1-u^2}}\,du\\\\ &=\arcsin(u)+C\\\\ &=\arcsin\left(\frac{\sqrt{x^2-1}}{x}\right)+C\\\\ &=-\arcsin(1/x)+C' \end{align}$$

where we used the identity $\arcsin(1/x)+\arcsin\left(\frac{\sqrt{x^2-1}}{x}\right)=(\pi/2)\text{sgn}(x)=\pi/2$ when $x>1$.

One can proceed similarly for the case in which $x<-1$.

Mark Viola
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HINT:

$$\dfrac1{x^2\sqrt{1-\dfrac1{x^2}}}=\dfrac{|x|}{x^2\sqrt{x^2-1}}$$

Set $\sqrt{x^2-1}=u\implies x^2=u^2+1$ and $x\ dx=u\ du$

lab bhattacharjee
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