I'm asked to show that the arc length of the curve $ \boldsymbol{\alpha}(t)= (e^{-t}\cos t,e^{-2t} \sin t)$ is finite. Now, clearly $$ \boldsymbol{\alpha}'(t)= (-e^{-t}(\cos t+\sin t),e^{-2t}(\cos t -2\sin t)) \implies\mid \mid \boldsymbol {\alpha'} (t) \mid \mid^2 = ( e^{-2t}(\cos t+\sin t)^2+e^{-4t}(\cos t -2\sin t)^2).$$
So I've reduced $\mid \mid \boldsymbol {\alpha'} (t) \mid \mid$ down to $$\mid \mid \boldsymbol {\alpha'} (t) \mid \mid=e^{-2t} \sqrt{ (1+3\sin^2 2t-2\sin2t+e^{2t}+e^{2t}\sin2t}$$ and equivalently $$\mid \mid \boldsymbol {\alpha'} (t) \mid \mid=\frac{ e^{-2t}\sqrt{5 + 2 e^{2t} - 3 \cos{2t} + -4\sin 2t + 2 e^{2t} \sin2t}}{\sqrt2}$$ but I can't see to find a method for integrating this that works. I've tried to integrate with both versions (and others) of $\mid \mid \boldsymbol {\alpha'} (t) \mid \mid$ but I get stuck midway or the integrating becomes too messy / repetitive. Consequently, I still can't seem to show that $$\int_0^\infty \mid \mid \boldsymbol {\alpha'} (u) \mid \mid du = \lim _{t\rightarrow\infty} \int_0^t \mid \mid \boldsymbol {\alpha'} (u) \mid \mid du <\infty.$$
I've considered showing that $\lim _{t\rightarrow\infty} \int_0^t \mid \mid \boldsymbol {\alpha'} (u) \mid \mid^2 du <\infty$ and then using this to imply the above but I can't find any theorems that allow me to do this. Does anyone have an idea as to a method (suitable for a third year university student) that I could use?
$$e^{-2t}(\cos t+\sin t)^2+e^{-4t}(\cos t -2\sin t)^2 <2e^{-2t}+5e^{-4t}$$
$$\int_0^\infty \sqrt{e^{-2t}(\cos t+\sin t)^2+e^{-4t}(\cos t -2\sin t)^2 \,dt }<\int_0^\infty \sqrt{2e^{-2t}+5e^{-4t}}\,dt$$ $$\int_0^\infty \sqrt{2e^{-2t}+5e^{-4t}}\,dt=\frac{\sqrt{7}}{2}+\frac{\sinh ^{-1}\left(\sqrt{\frac{5}{2}}\right)}{\sqrt{5}}\sim 1.87695$$