Find the curve length when $\alpha(t)=(a(t-\sin t),a(1-\cos t))$ where $a>0$ and $t\in [0,2\pi]$
So I am using: $$\int_{a}^{b}\|\alpha'(t)\|\,\mathrm dt$$
$\alpha'(t)=(a-a\cos t,a\sin t)$
$$\|\alpha'(t)\|=\sqrt{(a-a\cos t)^2+(a\sin t)^2}=\sqrt{a^2-2a^2 \cos t+a^2\cos^2t+a^2\sin^2t}=\sqrt{a^2-2a^2 \cos t+a^2(\cos^2t+\sin^2t)}=\sqrt{2a^2-2a^2 \cos t}=\sqrt{2a^2(1-\cos t)}=\sqrt{2a^2(2\sin^2(t/2))}=\sqrt{4a^2(\sin^2(t/2))}=2a\sin(t/2)$$
So we have $$\int_{0}^{2\pi}2a\sin(t/2)\,\mathrm dt=2a\int_{0}^{\pi}\sin(t/2)\,\mathrm dt=2a(-2\cos(t/2))|^{\pi}_{0}=4a$$
But the answer is $8a$
There's a typo in the integral calculation. $$\int_0^{2\pi}2a\sin\frac t2\,dt=2a\int_0^{\mathbf 2\pi}\sin\frac t2\,dt=2a\left[-2\cos\frac t2\right]^{2\pi}_0=8a$$