Integral of$\int_0^1 x\sqrt{2- \sqrt{1-x^2}}dx$ I have no idea how to do this, it seems so complex I do not know what to do.
$$\int_0^1 x\sqrt{2- \sqrt{1-x^2}}dx$$
I tried to do double trig identity substitution but that did not seem to work.
 A: Here is how I would do it, and for simplicity I would simply look at the indefinite integral.
First make the substitution $u = x^2$ so that $du = 2xdx$. We get: $$\frac{1}{2} \int \sqrt{2-\sqrt{1-u}} du$$ Then, make the substitution $v = 1-u$ so that $dv = -du$. We get: $$-\frac{1}{2} \int \sqrt{2 - \sqrt{v}} dv$$ Then make the substitution $w = \sqrt{v}$ so that $dw = \frac{1}{2\sqrt{v}} dv$ meaning that $dv = 2w \text{ } dw$. So we get: $$-\int \sqrt{2-w} \text{ } w \text{ } dw$$ Now make the substitution $s = 2-w$ so that $ds = -dw$ to get: $$-\int \sqrt{s}(s-2) ds$$ The rest should be straightforward.
A: Since you tried to use a trig identity, I'll use one in this solution. Let $x = \sin \theta$ so that $1 - x^2 = 1 - \sin^2 \theta = \cos^2 \theta$, and $\mathrm d x = \cos \theta \mathrm{d}\theta$. Our integral becomes:
$$ \int_0^{\frac \pi 2} \sin \theta \cos \theta \sqrt{2 - \cos\theta} \mathrm d \theta.$$
Now set $u = \cos\theta$ to give $\mathrm d \mathrm u = - \sin \theta \mathrm d \theta$. Our integral becomes
$$\int_1^0 - u\sqrt{2-u} \mathrm \ \mathrm d u.$$
Can you solve that?
