## Calculus: Early Transcendentals 8th Edition

$\displaystyle \frac{2}{15}(8\sqrt{2}-7)\approx 0.57516$
Substituting so that so $u^{2}=1-x^{2},\ \quad (u=\sqrt{1-x^{2}})$ $2udu=-2xdx$, $xdx=-udu,$ the integral becomes simpler bounds:$\left[\begin{array}{l} x=0\rightarrow u=1\\ x=1\rightarrow u=0 \end{array}\right]$: $I=\displaystyle \int_{0}^{1}x\sqrt{2-\sqrt{1-x^{2}}}dx=\int_{1}^{0}\sqrt{2-u}(-udu)$ $\left[\begin{array}{lll} v=\sqrt{2-u} & v^{2}=2-u & u=2-v^{2}\\ & 2vdv=-du & \\ & & \end{array}\right] \left[\begin{array}{l} u=0\rightarrow v=\sqrt{2}\\ u=1\rightarrow v=1 \end{array}\right]$ $I=\displaystyle \int_{1}^{\sqrt{2}}v(2-v^{2})(2vdv)$ $=\displaystyle \int_{1}^{\sqrt{2}}(4v^{2}-2v^{4})dv$ ... both are table integrals type 1 $=[\displaystyle \frac{4}{3}v^{3}-\frac{2}{5}v^{5}]_{1}^{\sqrt{2}}$ $=(\displaystyle \frac{4}{3}2^{3/2}-\frac{2}{5}2^{5/2})-(\frac{4}{3}-\frac{2}{5})$ $=(\displaystyle \frac{8}{3}2^{1/2}-\frac{8}{5}2^{1/2})-(\frac{14}{15})$ $=\displaystyle \frac{16}{15}2^{1/2}-\frac{14}{15}$ $=\displaystyle \frac{2}{15}(8\sqrt{2}-7)\approx 0.57516$