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

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.

• for the indefinite integral i see not many chances. Maybe if there are some limits – tired Feb 11 '15 at 18:27
• 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: $$-L \int{\frac{tan(u)}{u} du}$$ – Upax Feb 11 '15 at 18:35

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.