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For a physical problem I have to solve $\sqrt{\frac{m}{2E}}\int_0^{2\pi /a}\frac{1}{(1-\frac{U}{E} \tan^2(ax))^{1/2}}dx $

I already tried substituting $1-\frac{U}{E}\tan^2(ax)$ and $\frac{U}{E}\tan^2(ax)$ since $\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin(x)$ but my problem is that $dx$ changes to something with $\cos^2(ax)$, thus making the integral not easier.

Anyone got a hint?

EDIT: The physical problem is to calculate the oscillating period given an potential $V(x) = U \tan^2(ax) $ by using conservation of energy. Here's what I did so far:

$E_{kin} + V(x) = E $

$=> \frac12 m (\frac{dx}{dt})^2 = E-V(x)$

$=>\int_{0}^{T}dt = \sqrt{\frac{m}{2E}}\int_{0}^{2\pi/a} \frac{dx}{(1-U/E \tan^2(ax))^{1/2}}$

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  • $\begingroup$ Replace the $\tan$ by a new variable. Since the derivative of the $\arctan$ is rational, you get an integral of the form $\int R(x,\sqrt{az^2+bz+c})$ where $R$ is rational. Then use Euler substirutions. $\endgroup$ – Pp.. Jan 31 '15 at 19:29
  • $\begingroup$ Can you go more in detail with that? Because it doesn't solve my problem that $dx$ changes to something with $cos^2(ax)$ $\endgroup$ – Christian Jan 31 '15 at 19:58
  • $\begingroup$ Are you really sure the period is given by such an integral? What happens when $\tan^2(\alpha x)>\frac{E}{U}$? The square root is not defined in such a case. $\endgroup$ – Jack D'Aurizio Jan 31 '15 at 20:04
  • $\begingroup$ I will edit the original task $\endgroup$ – Christian Jan 31 '15 at 20:07
  • $\begingroup$ I rewrote my answer a few days ago. I'll like you take a look at that and ignores the wrong comment at it. $\endgroup$ – Felix Marin Feb 10 '15 at 21:49
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Let me just work with $$I:=\int\frac{1}{\sqrt{1-b\tan^2(ax)}}dx$$

Once you get this primitive you know how to compute your definite integral.

Let's put $y=\tan(ax)$. Then $x=\frac{1}{a}\arctan(y)$, and $dx=\frac{1}{a}\frac{1}{1+y^2}dy$.

Then $$I=\frac{1}{a}\int\frac{1}{\sqrt{1-by^2}}\frac{1}{1+y^2}dy.$$

We can use an Euler substitution such that $\sqrt{1-by^2}=yz-1$ (the second type).

Then $1-by^2=y^2z^2-2yz+1$. From where $0=yz^2-2z +by$. We get then that $$\begin{align}y&=\frac{2z}{z^2+b}\\dy&=\frac{2b-2z^2}{(z^2+b)^2}dz\\\sqrt{1-by^2}&=\frac{2z}{z^2+b}\cdot z-1\end{align}$$

Putting this into the integral we get $$I=\frac{1}{a}\int\frac{1}{\frac{2z}{z^2+b}\cdot z-1}\frac{1}{1+\left(\frac{2z}{z^2+b}\right)^2}\frac{2b-2z^2}{(z^2+b)^2}dz$$

Observe how this is only the integral of a rational function. You can use simple fraction decomposition to compute it. Personally, I would give this last to a computer to do it for me.

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  • $\begingroup$ How do you get from $\sqrt{1-by^2}$ to $\frac{2z}{z^2+b}z-1$ ? $\endgroup$ – Christian Feb 1 '15 at 9:24
  • $\begingroup$ Okay, I see by using $\sqrt{1-by^2}=yz-1$, I got stuck with it because I wanted to substitute $y$ in $\sqrt{1-by^2}$ with $y=\frac{2z}{z^2+b}$ $\endgroup$ – Christian Feb 1 '15 at 9:33
  • $\begingroup$ @Christian Yes. I tried leaving it without simplification so you can see where it is coming from. Where were you getting that $\cos^2(ax)$ that you were saying? $\endgroup$ – Pp.. Feb 1 '15 at 15:01
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ I guess the OP is considering an application of the BS-quantization rule. Indeed the original question is not right at all. You should consider an interval, along the $\ds{x}$-axis, where the kinetic energy is $\ds{\geq 0}$. It defines the 'returning points' $\ds{\braces{x}}$ which are given by $\ds{U\tan^{2}\pars{ax}=E}$.

Hereafter we'll assume that $\ds{U > 0}$ such that the energy $\ds{E}$ is a positive quantity $\ds{\pars{~E > 0~}}$. For simplicity we'll assume the 'problem' is 'confined' to symmetrical returning points which are given by $\ds{\pm\,\tilde{x}}$ where $\ds{\tilde{x} = \frac{1}{a}\,\arctan\pars{\root{\frac{E}{U}}}}$.

There isn't any essential difficult whenever we consider a more general situation. We are lead to evaluate: \begin{align}&\color{#66f}{\large% \root{\frac{m}{2E}}\int_{-\tilde{x}}^{\tilde{x}} \frac{\dd x}{\bracks{1 -U\tan^{2}\pars{ax}/E}^{1/2}}} =\frac{2}{a}\root{\frac{m}{2E}} \ \overbrace{% \int_{0}^{a\tilde{x}}\frac{\dd x}{\bracks{1 -U\tan^{2}\pars{x}/E}^{1/2}}} ^{\ds{\dsc{\tan\pars{x}}\equiv\dsc{t}\ \imp\ \dsc{x}=\dsc{\arctan\pars{t}}}} \\[5mm]&=\frac{2}{a}\root{\frac{m}{2E}}\ \overbrace{% \int_{0}^{\root{E/U}} \frac{\dd t}{\bracks{1 -Ut^{2}/E}^{1/2}\pars{t^{2} + 1}}} ^{\ds{\dsc{t}\ \mapsto\ \dsc{\frac{1}{t}}}} \\[5mm]&=\frac{2}{a}\root{\frac{m}{2E}}\ \overbrace{% \int_{\root{U/E}}^{\infty} \frac{t\,\dd t}{\bracks{t^{2} -U/E}^{1/2}\pars{t^{2} + 1}}} ^{\ds{\dsc{t^{2}}\ \mapsto\ \dsc{t}}} =\frac{1}{a}\root{\frac{m}{2E}}\ \overbrace{\int_{U/E}^{\infty} \frac{\dd t}{\bracks{t -U/E}^{1/2}\pars{t + 1}}} ^{\ds{\dsc{\pars{t - U/E}^{1/2}}\ \mapsto\ \dsc{t}}} \\[5mm]&=\frac{2}{a}\root{\frac{m}{2E}}\int_{0}^{\infty} \frac{\dd t}{t^{2} + U/E + 1} =\frac{2}{a}\root{\frac{m}{2E}}\frac{1}{\root{1 + U/E}}\ \overbrace{% \int_{0}^{\infty} \frac{\dd t}{t^{2} + 1}}^{\dsc{\frac{\pi}{2}}} \end{align}


Finally, \begin{align}&\color{#66f}{\large% \root{\frac{m}{2E}}\int_{-\tilde{x}}^{\tilde{x}} \frac{\dd x}{\bracks{1 -U\tan^{2}\pars{ax}/E}^{1/2}}} =\color{#66f}{\large\frac{\pi}{a}\root{\frac{m}{2\pars{E + U}}}} \end{align} This result together with the BS-quantization rules provides an expression for the energy $\ds{E}$.

I suggest that, next time, you post this kind of question in Physics StackExchange.

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  • $\begingroup$ This is plagued by errors. A hint that something went astray is the square root of a negative number at the end (but already the second line is wrong). $\endgroup$ – Did Feb 2 '15 at 4:12

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