Integral equation

$\begin{aligned} \int_0^\frac{\pi}{a} e^{\tan x}(\tan^3 x + \tan x) \space dx & = 1 \\ \int_0^\frac{\pi}{a} e^{\tan x}\tan x(\tan^2 x + 1) \space dx & = 1 \\ \int_0^\frac{\pi}{a} e^{\tan x}\tan x \sec^2 x \space dx & = 1 \quad \quad \small \color{#3D99F6}{\text{Let } t = \tan x \quad \Rightarrow dt = \sec^2 x \space dx} \\ \Rightarrow \int_0^{\tan \frac{\pi}{a}} \color{#D61F06}t \color{#3D99F6}{e^{t} \space dt} & = 1 \quad \quad \small \color{#3D99F6}{\text{By integration by parts: } \color{#D61F06} {u = t} \quad \color{#3D99F6} dv = e^t \space dt} \\ \left[te^t\right]_0^{\tan \frac{\pi}{a}} - \int_0^{\tan \frac{\pi}{a}} e^{t} \space dt & = 1 \\ e^{\tan \frac{\pi}{a}} \tan \frac{\pi}{a}- \left[e^t\right]_0^{\tan \frac{\pi}{a}} & = 1 \\ e^{\tan \frac{\pi}{a}} \tan \frac{\pi}{a}- e^{\tan \frac{\pi}{a}} + 1 & = 1 \\ e^{\tan \frac{\pi}{a}} \tan \frac{\pi}{a} & = e^{\tan \frac{\pi}{a}} \\ \Rightarrow \tan \frac{\pi}{a} & = 1 \\ \Rightarrow a & = \boxed{4} \end{aligned}$

$\int_0^\frac{\pi}{4} e^{\tan x}((\tan x)^3 + \tan x) dx =\int_0^\frac{\pi}{4} e^{\tan x}\tan x((\tan x)^2 + 1) dx =$ $\left[e^{tan x} \cdot \tan x\right]^\frac{\pi}{4}_0 - \int_0^\frac{\pi}{4} e^{\tan x}((\tan x)^2 + 1) dx$ $= e - \left[e^{\tan x} \right]^\frac{\pi}{4}_0 = e - ( e - 1) =1$

Question: are there more possibilities for a?