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 $

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

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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.