Solving an integral (with substitution?) 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 $ 
$=> \frac12 m (\frac{dx}{dt})^2 = E-V(x)$
$=>\int_{0}^{T}dt = \sqrt{\frac{m}{2E}}\int_{0}^{2\pi/a} \frac{dx}{(1-U/E \tan^2(ax))^{1/2}}$
 A: 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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
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 \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\,}\,}
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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.

