How can I evaluate $$\int_L{\frac{z^2}{x^2+y^2}}$$, where:
L: $$x=\cos{t};$$ $$y=a\cdot\sin{t};$$ $$z=a\cdot t$$ $$0\leq t \leq 2\pi$$
So far I tried it this way:
$$I=\int_L{\frac{z^2}{x^2+y^2}}ds=\int_0^{2\pi}{(\frac{a^2t^2}{\cos^2{t}+a^2\cdot\sin^2{t}}\sqrt{(\frac{d(\cos{t})}{dt})^2 + (\frac{d(a\cdot \sin{t})}{dt})^2 +(\frac{d(a\cdot t)}{dt})^2}})dt=\int_0^{2\pi}{(\frac{a^2t^2}{\cos^2{t}+a^2\cdot\sin^2{t}}\sqrt{\sin^2{t} + a^2\cdot\cos^2{t} +a^2}})dt$$
and that is where I am getting stuck as I can't reduce of simplify this expression. How can this problem be solved?