Find the arclength of the curve defined by $$x=\cos^2(t)$$$$y=\cos(t)$$ from $0$ to $4\pi$.
I know using the formula that the arclength is given by $$\int_{\alpha}^{\beta}\sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2+\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\ \mathrm{d}t$$ and that gives me the integral $$\int_0^{4\pi} \sqrt{4\cos^2(t)\sin^2(t)+\sin^2(t)}\mathrm{d}t$$ which yields $$\int_0^{4\pi}\sin(t)\sqrt{4\cos^2(t)+1}\ \mathrm{d}t$$ which becomes $$\int_{1}^{1}\sqrt{4u^2+1}\ \mathrm{d}u=0$$ Through u-substitution of $u=\cos(t)$. The arclength cannot be $0$. What am I doing wrong?
Your substitution has to be a bijection, or else you lose bits of the integral, as you have noticed: draw a picture of the substitution, and notice that you've lost bits. Also, as @GudsonChou points out, $\sqrt{\sin^2{t}}$ is only equal to $\sin{t}$ half the time.
If you draw the curve, you should notice that it's both symmetric about $t=\pi$ and covers the same path twice. Hence it is sufficient to compute $$ 4\int_0^{\pi} \sin{t} \sqrt{4\cos^2{t}+1} \, dt = 4\int_0^1 \sqrt{4u^2+1} \, du, $$ which you can of course do using a couple of possible substitutions, trigonometric or hyperbolic.