Answer
$$\displaystyle\int_0^1{x\sqrt{2-\sqrt{1-x^2}}dx}=\displaystyle\frac{16}{15}\sqrt{2}-\frac{14}{15}$$
Work Step by Step
$I=\displaystyle\int_0^1{x\sqrt{2-\sqrt{1-x^2}}dx}$
$\displaystyle{\left[\begin{matrix}
u=\displaystyle\sqrt{1-x^2}& dx=\displaystyle\frac{du\sqrt{1-x^2}}{x}\\
\end{matrix}\right]}$
$I=\displaystyle\int_1^0{x\sqrt{2-u}\times \frac{du\sqrt{1-x^2}}{x}}\\
I=\displaystyle\int_1^0{u\sqrt{2-u}\ du}$
$\displaystyle{\left[\begin{matrix}
z=\sqrt{2-u}& du=2\sqrt{2-u}\ dz\\
\end{matrix}\right]}$
$I=\displaystyle\int_1^{\sqrt{2}}{uz\times 2\sqrt{2-u}\ dz}\\
I=\displaystyle2\int_1^{\sqrt{2}}{\left( 2-z^2 \right) z^2\ dz}\\
I=\displaystyle2\int_1^{\sqrt{2}}{2z^2-z^4\,\,dz}\\
I=\displaystyle2\left[ \frac{2}{3}z^3-\frac{1}{5}z^5 \right] _{1}^{\sqrt{2}}\\
I=\displaystyle\frac{16}{15}\sqrt{2}-\frac{14}{15}$
