Calculate $\int_{1}^{-1} |z-1|^2 dz$ along the upper half of the circle $|z|=1$

Question

Calculate $$\int_{1}^{-1} |z-1|^2 dz$$ along the upper half of the circle $|z|=1$

This was my attempted solution

We use the parameterization $z = e^{it} = \cos t + i\sin t$ then $|z-1|^2 = \sqrt{2}\sqrt{1-\cos t}$ and then we get $$\int_{-1}^{1} |z-1|^2dz = \sqrt{2}i \int_{0}^\pi \cos t \sqrt{1- \cos t} + i \sin t\sqrt{1- \cos t}dt$$

Now using a simple $u$-substitution I can calculate $$\int_{0}^{\pi} i \sin t\sqrt{1- \cos t}dt = \frac{2^{\frac{5}{2}}i}{3}$$ however I'm not sure how to calculate $\int_{0}^\pi \cos t \sqrt{1- \cos t}$ using any $u$-substitution. I can calculate it numerically using WolframAlpha, but that defeats the purpose as this is an exam testable question in a course I'm taking where we don't use any calculators.

Is what I've done so far correct? If so how can I calculate the above integral by hand. Are there any "nicer" parameterizations that I could use to avoid this problem?

Answer 1

Hint. We have $$ |z-1|^2=|z|^2+1-2\mathrm{Re}\,z $$ So if $\,z=\mathrm{e}^{it}$, then $$ |\mathrm{e}^{it}-1|^2=2-2\cos t $$ and hence, integrating on the upper half unit disc we obtain $$ \int_{1}^{-1} |z-1|^2\,dz=\int_{0}^\pi (2-2\cos t)\,i\mathrm{e}^{it}\,dt $$ since $dz=i\mathrm{e}^{it}\,dt$.