$$ \int \frac{x^{n+2}\ dx}{\sqrt{(1-x^2)}} = \frac{n+1}{n+2} \int \frac{x^n \ dx}{\sqrt{(1-x^2)}} - \frac{x^{n+1}}{n+2} \sqrt{1-x^2} $$
It is said that it can be solved by integration by parts. I've spend last few hours checking almost every possibility, as $ u = x^{n+1}, x^{n+2} ...\ dv = \frac{dx}{\sqrt{(1-x^2)}x} $ etc. I've surely wasted half of the forest (poor Euler) writing such monster equations with five or more integrations by parts inside. I would be thankful for any help.
I'm a little bit afraid that solution is really easy but probably weariness wouldn't help me.
(source: http://www.math.uiuc.edu/~reznick/sandifer.pdf page 10)
It can be done by parts, without the trig. Let $I_n=\int \frac{x^n \ dx}{\sqrt{1-x^2}},$ and consider the integral $$J_n=\int \frac{x^n(1-x^2)}{\sqrt{1-x^2}}.$$ On the one hand it is $I_n-I_{n+2}$ by expanding the numerator. On the other hand, simplifying the integrand of $J_n$ to $x^n \sqrt{1-x^2},$ we can integrate $J_n$ by parts with $u=\sqrt{1-x^2}$ and $dv=x^n\ dx.$ This gives $$J_n = \sqrt{1-x^2}\frac{x^{n+1}}{n+1}+\frac{1}{n+1}I_{n+2}.$$ If we now equate the two versions of $J_n$ and solve for $I_{n+2}$ we arrive at the reduction formula at the top of the question.