# 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.

-
For future reference, searches like "integral of x sqrt(2 - sqrt(1 - x^2)) dx" on wolframalpha.com often work. (Hit "show steps...") – leslie townes Jun 5 '12 at 0:02
There are often multiple ways to solve these integrals, as demonstrated below. You just have to experiment with different substitutions. Also bear in mind that these integrals are designed to be solved, so there is usually a simple trick. If you see a $1-x^2$, we can try sin or cos because that will get rid of a square term. If you have linear terms you can try making a substitution by changing the term that is in an awkward place, for example under a square root sign. – Paul Slevin Jun 5 '12 at 7:09

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?

-
All I used here was that $\sin ^2 \theta + \cos ^2 \theta = 1$. That is the most important trig identity for these types of integrals (IMO). – Paul Slevin Jun 5 '12 at 0:07
I do not understand what you did to change the bounds. I can not solve the problem from there. – user138246 Jun 5 '12 at 0:55
sin(π/2) = 1 , sin(0) = 0, cos(π/2) = 0, and cos(0) = 1 – hungr Jun 5 '12 at 3:06
Follow the rules given in your notes. hungr has explained the bounds, and in order to solve it, simply make a linear substitution to get rid of the 2-u from under the square root sign. – Paul Slevin Jun 5 '12 at 6:52
To add more detail to the bounds part, because I changed variable by letting $x = \sin\theta$ and $x$ was already in my equation, I had to apply $\operatorname{arcsin}$ to the bounds. In the 2nd substitution, I just applied $\cos$ to the bounds in the standard way. – Paul Slevin Jun 5 '12 at 6:58

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.

-