Integrate $\int x \sqrt{2 - \sqrt{1-x^2}}dx $ it seems that integration by parts with some relation to substitution...

$$
\int x \sqrt{2-\sqrt{1-x^2}}
= \frac{2}{5} \sqrt{2-\sqrt{1-x^2}} \cdot \sqrt{1-x^2}+\frac{8}{15}\sqrt{2-\sqrt{1-x^2}}+c
$$

How can I get that?
 A: Let $x=\sqrt{1-y^2}$, then then integral becomes
$$-\int dy \, y \, \sqrt{2-y}$$
Integrate by parts:
$$\frac{2}{3}y (2-y)^{3/2} + \frac{2}{3} \int dy \, (2-y)^{3/2}$$
which is
$$\frac{2}{3}y (2-y)^{3/2} - \frac{4}{15} (2-y)^{5/2} + C$$
Therfore
$$\int dx \, x \, \sqrt{2-\sqrt{1-x^2}} = \frac{2}{3} \sqrt{1-x^2} \left ( 2-\sqrt{1-x^2}\right)^{3/2} - \frac{4}{15}\left ( 2-\sqrt{1-x^2}\right)^{5/2}+C $$
A: Let, $1 - x^2 = u^2 \implies - 2x dx = 2u  du $, this gives your integral value, 
$ \displaystyle - \int u \sqrt{2 - u}  \; \mathrm du $. Now try using integration by parts.
A: You don't need any integrations by parts for this. As others have suggested, if you substitute $u^2 = 1 -x^2$ and $2u\,du = -2x\,dx$ the integral becomes
$$-\int u\sqrt{2 - u}\,du$$
Now do another substitution $v = 2 - u$ and the integral becomes
$$\int (2-v)\sqrt{v}\,dv$$
$$= \int 2v^{1 \over 2} - v^{3 \over 2}\,dv$$
$$= {4 \over 3} v^{3 \over 2} - {2 \over 5} v^{5 \over 2} + C$$
Substituting back $v = 2 - u = 2 - \sqrt{1 - x^2}$ this becomes
$${4 \over 3} (2 - \sqrt{1 - x^2})^{3 \over 2} - {2 \over 5} (2 - \sqrt{1 - x^2})^{5 \over 2} + C$$
This is the final answer.
A: Welcome to StackExchange! You may use LaTeX-Formulas by using dollar signs \$ x^2 \$ $\longrightarrow{}$ $x^2$.
As a first step I used the substitution $u:=\sqrt{2-x^2}$ which implies $ \frac{\mathrm{d}u}{\mathrm{d}x}=\frac{2x}{2\sqrt{1-x^2}}=\frac{x}{u}$. So we have to solve $\int \sqrt{2-u}\cdot u\ \mathrm{d}u$. This can be done by using partial integration (http://en.wikipedia.org/wiki/Integration_by_parts, use $v'=\sqrt{2-u}$), which yields $-2/15 (2-u)^ {3/2} (4+3 u)+c$. This should be equivalent to your result, if you replace $u$ by $\sqrt{1-x^2}$.
