Integration by parts $ \int \sqrt{x-x^2} dx $ I have to integrate $$ \int \sqrt{x-x^2} dx $$ 
The answer on my textbook is $ \frac 14 \left( \arcsin(\sqrt x)-(1-2x)\sqrt{x-x^2} \right) $ but I want to solve this by myself so can you please give me a clue ? :) Thank you.
Hint: Complete the square:
$$\int\sqrt{x-x^2}dx=\int\sqrt{-x^2+2\frac x2-\frac14+\frac14}dx=\int\sqrt{-\left(x-\frac 12\right)^2+\frac14}dx$$
Set $u=x-\frac 12$ to obtain
$$\int\sqrt{-u^2+\frac14}du$$
I think you now how to deal with this one.
for calculation of Integral $\displaystyle \int \sqrt{a^2-x^2}dx$

We will Use Integration by parts method

Let $\displaystyle \mathbb{I} = \int \sqrt{a^2-x^2}.xdx = \sqrt{a^2-x^2}.x-\int\frac{1}{2\sqrt{a^2-x^2}}.-2x.xdx$

$\displaystyle \mathbb{I}= x.\sqrt{a^2-x^2}-\int\frac{(a^2-x^2)-a^2}{\sqrt{a^2-x^2}}dx$

$\displaystyle \mathbb{I} = x.\sqrt{a^2-x^2}-\mathbb{I}+a^2.\sin^{-1}\left(\frac{x}{a}\right)$

So $\displaystyle \mathbb{I}=\sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}.\sin^{-1}\left(\frac{x}{a}\right)+\mathbb{C}$