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I'm stuck on an integral

$$\int \tan \left( a_0 \, e^{(-x/L)} \right) \,dx $$

where $a_0$ is a constant.

I've tried doing a substitution with $ u = a_0e^{x/L}$ but it gave me

$$\int \frac{\tan(u)}{u}du $$ which Wolfram Alpha didn't like.

Any help would be appreciated.

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  • $\begingroup$ for the indefinite integral i see not many chances. Maybe if there are some limits $\endgroup$ – tired Feb 11 '15 at 18:27
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    $\begingroup$ If $u=a_0 e^{x/L}$ then $du= \frac{1}{L} u dx$. However, please note that inside your integral you have $u=a_0 e^{-{x/L}}$ so I suppose you use these substitution since the argument of your tangent is just u. In this case your integral becomes: \begin{equation} -L \int{\frac{tan(u)}{u} du} \end{equation} $\endgroup$ – Upax Feb 11 '15 at 18:35
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Just like other trigonometric integrals, it does not possess a closed form expression. See also inverse tangent integral. This can be proven using either Liouville's theorem or the Risch algorithm. Alternately, expand $\tan(x)$ into its well-known Taylor series, then reverse the order of summation and integration.

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