Find the Arc length of the parametric curve

Question

$$x=6t-6sint$$
$$y=6-6cost$$
Find the arc length of the parametric curve
$$Arc length = \int_{0}^{2\pi} \sqrt{(6-6cost)^2+(6sint)^2}dt\\
=\int_{0}^{2\pi} \sqrt{36-72cost+36cos^2t+36sin^2t}dt \\
=\int_{0}^{2\pi} 6 \sqrt{1-2cost+cos^2t+sin^2t}dt\\
=\int_{0}^{2\pi} 6\sqrt{2-2cost}dt\\
=\int_{0}^{2\pi} 6\sqrt{2}\sqrt{1-cost}dt\\
= 6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}dt\\
=6\sqrt{2}\int_{0}^{2\pi}\sqrt{1-cost}\frac{\sqrt{(1+cost)}}{\sqrt{(1+cost)}} dt\\
=6\sqrt{2}\int_{0}^{2\pi}\frac{\sqrt{(1-cos^2t)}}{\sqrt{(1+cost)}} dt\\
=6\sqrt{2}\int_{0}^{2\pi}\frac{sint}{\sqrt{(1+cost)}} dt\\$$
Let $$u =1+cost$$ $$du=-sint$$
$$=-6\sqrt{2}\int_{0}^{2\pi}\frac{1}{\sqrt{u}}\\
=-12\sqrt{2}[u^\frac{1}{2}]\\
=0$$
But the answer is 48.

Answer 1

Leaving the coefficient $6$ on the side, we have

$$s=\int_0^{2\pi}\sqrt{(1-\cos t)^2+(\sin t)^2}\,dt=\int_0^{2\pi}\sqrt{2-2\cos t}\,dt=2\int_0^{2\pi}\left|\sin\frac t2\right|\,dt.$$