Evaulate line integral $\int_k y^2 ds$ for $k: x=2(t-\sin{t}),\; y=2(1-\cos{t}),\; t\in[0,2\pi]$.
I got into a point where I have the following:
$8\sqrt{2}\int_0^{2\pi}(1-\cos{t})^2\sqrt{1-\cos{t}}\;\mathrm{d}t$
unable to solve it. The result should be $\frac{2048}{15}$.
Note that\begin{align}1-\cos t&=\cos^2\left(\frac t2\right)+\sin^2\left(\frac t2\right)-\left(\cos^2\left(\frac t2\right)-\sin^2\left(\frac t2\right)\right)\\&=2\sin^2\left(\frac t2\right)\end{align}and that therefore\begin{align}\int_0^{2\pi}(1-\cos t)^2\sqrt{1-\cos t}\,\mathrm dt&=\int_0^{2\pi}4\sin^4\left(\frac t2\right)\sqrt2\sin\left(\frac t2\right)\,\mathrm dt\\&=4\sqrt2\int_0^{2\pi}\left(1-\cos^2\left(\frac t2\right)\right)^2\sin\left(\frac t2\right)\,\mathrm dt.\end{align}Now, you can compute this integral doing the substitution $\cos\left(\frac t2\right)=u$ and $-\frac12\sin\left(\frac t2\right)\,\mathrm dt=\mathrm du$.