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Hi guys I'm trying to solve the following integral that I've found in a paper:

$\int_{-l}^{l}$$\frac{\delta x}{((\frac{a-x}{r})^{2}+1)^{0.5}}$

The authors report the following result:

$sinh^{-1}(\frac{l-a}{r})+sinh^{-1}(\frac{l+a}{r})$

Any idea on how the steps required to obtain this result?

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    $\begingroup$ Let $\frac{a-x}{r}=\sinh t$. We end up wanting $\int k\,dt$ for a certain constant! $\endgroup$ – André Nicolas Aug 7 '14 at 14:22
  • $\begingroup$ General idea: sum of squares suggests a trigonometric substitution. (In this case, when you are outside the range of the natural trig substitution, you can use a hyperbolic substitution instead.) $\endgroup$ – GEdgar Aug 7 '14 at 14:23
  • $\begingroup$ What do you mean by hyperbolic substitution? A change of variable? $\endgroup$ – SSC Napoli Aug 7 '14 at 14:39
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folowing the hint: $$\int_{-l}^{l}\frac{dx}{\sqrt{\left(\frac{a-x}{r}\right)^2+1}}\\ \sinh t=\frac{a-x}{r}\iff r\sinh t=a-x\\ r\cosh t dt=-dx\iff-r\cosh t=dx\\ -\sinh^2t+\cosh^2t=1\Rightarrow\cosh^2t=1+\sinh^2t\\ \int\frac{dx}{\sqrt{\left(\frac{a-x}{r}\right)^2+1}}=-r\int\frac{\cosh tdt}{\sqrt{\sinh^2t+1}}=-r\int\frac{\cosh tdt}{\sqrt{\cosh^2 t}}=\\ -r\int\frac{\cosh tdt}{\cosh t}=-r\int dt=-rt+C=-r~\sinh^{-1}\left(\frac{a-x}{r}\right)+C\\ \int_{-l}^{l}\frac{dx}{\sqrt{\left(\frac{a-x}{r}\right)^2+1}}=\left.-r~\sinh^{-1}\left(\frac{a-x}{r}\right)\right|^{l}_{-l}=\\ -r~\sinh^{-1}\left(\frac{a-l}{r}\right)+r~\sinh^{-1}\left(\frac{a+l}{r}\right)$$


then lets proof that $\displaystyle\int\frac{dx}{\sqrt{\left(\frac{a-x}{r}\right)^2+1}}=-r~\sinh^{-1}\left(\frac{a-x}{r}\right)+C$

by fundamental theorem of calculus: $\int f(x)dx=F(x)\iff F'(x)=f(x)$

then

$\frac{d}{dx}\left[-r~\sinh^{-1}\left(\frac{a-x}{r}\right)+C\right]=\frac{d}{dx}-r~\sinh^{-1}\left(\frac{a-x}{r}\right)=-r\frac{d}{dx}\sinh^{-1}\left(\frac{a-x}{r}\right)$

using the fact that $\frac{d}{dx}\sinh^{-1}u=\frac{u'}{\sqrt{1+u^2}}$, then

$-r\frac{d}{dx}\sinh^{-1}\left(\frac{a-x}{r}\right)=-r\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}\frac{d}{dx}\frac{a-x}{r}=-r\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}\cdot-\frac{1}{r}=\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}$

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  • $\begingroup$ Thank you for the answer, however in this result the inverse hyperbolic functions are multiplied by $r$ while in the result that the authors report they are not... $\endgroup$ – SSC Napoli Aug 7 '14 at 15:02
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    $\begingroup$ @user3810266 cand's answer is correct. The result your authors reported is not. $\endgroup$ – David H Aug 7 '14 at 15:22
  • $\begingroup$ Yeah i just recognized that there is a $\frac{1}{r}$ outsider the integral that i forgot to report $\endgroup$ – SSC Napoli Aug 7 '14 at 15:46

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