How do I find the Integral of $\sqrt{r^2-x^2}$? How can I find the integral of the following function using polar coordinates ?

$$f(x)=\sqrt{r^2-x^2}$$ 

Thanks!

$\displaystyle\int \sqrt{r^2-x^2}dx$
Let be $\;x=r.\sin\alpha$   or   $\quad x=r.\cos\alpha$,
Let be $\;x=r.\sin\alpha$,
and $\quad dx=r.\cos \alpha \;d\alpha$
Integral be,
$\displaystyle\int \sqrt{r^2-x^2}dx=\displaystyle\int r.\sqrt{1-\sin^2\alpha}\;.r.\cos \alpha \;d\alpha=\displaystyle\int r^2.\cos^2\alpha\; d\alpha$
And use this equation,$\quad \cos2\alpha=2\cos^2\alpha-1\longrightarrow \cos^2\alpha=\dfrac{\cos 2\alpha+1}{2}$
Then,
$=\displaystyle\int r^2.\cos^2\alpha\; d\alpha=\displaystyle\int r^2.\dfrac{\cos 2\alpha+1}{2}\; d\alpha=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C$
From here ,
$\boxed{\;x=r.\sin\alpha}$
Answer'll be,
$\boxed{\boxed{\displaystyle\int \sqrt{r^2-x^2}dx=\dfrac{r^2}{2}\left[\dfrac{\sin2\alpha}{2}+\alpha\right]+C=\dfrac{r^2}{2}\left[\dfrac{2x\sqrt{r^2-x^2}}{r^2}+\arcsin\left(\dfrac{x}{r}\right)\right]}}$
