I want to find: $$\int\limits^{2\pi}_0\sin\frac{\theta}{2}\cdot\hat{r}d\theta$$ while $\hat r$ is as usually used in Physics (i.e. for $\theta=0.5\pi$ you get $\hat r=\hat y$, and for $\theta=\pi$ you get $\hat r=-\hat x$).
If I understand correctly, because $\text{sin}\frac{\theta}{2}=\text{sin}\left(\pi-\frac{\theta}{2}\right)=\text{sin}\left(\frac{2\pi-\theta}{2}\right)$, it holds that the $\hat x$ components of $\text{sin}\frac{\theta}{2}\cdot\hat{r}$ are identical for $\theta=t,2\pi-t$, while the $\hat y$ components are equal but opposite.
Therefore:
$$\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\cdot\hat{r}d\theta=\left(\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\cdot\hat{r}d\theta\hat{x}\right)\hat{x}=\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\text{cos}\theta d\theta\hat{x}$$
Now we have: $$\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\text{cos}\theta d\theta\hat{x}=\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\left(2\text{cos}^{2}\frac{\theta}{2}-1\right)d\theta\hat{x}$$ I managed to solve the last integral, and thus got: $$\overset{2\pi}{\underset{0}{\int}}\text{sin}\frac{\theta}{2}\cdot\hat{r}d\theta=-\frac{4}{3}\hat{x}$$
My question is:
Is my solution the simplest one, or have I missed some better way to solve the problem?