Definite integral evaluation, how to solve it? Hi guys I'm trying to solve the following integral that I've found in a paper:
$\int_{-l}^{l}$$\frac{\delta x}{((\frac{a-x}{r})^{2}+1)^{0.5}}$
The authors report the following result:
$sinh^{-1}(\frac{l-a}{r})+sinh^{-1}(\frac{l+a}{r})$
Any idea on how the steps required to obtain this result?
 A: folowing the hint:
$$\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
$-r\frac{d}{dx}\sinh^{-1}\left(\frac{a-x}{r}\right)=-r\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}\frac{d}{dx}\frac{a-x}{r}=-r\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}\cdot-\frac{1}{r}=\frac{1}{\sqrt{1+\left(\frac{a-x}{r}\right)^2}}$
