# Help on the following indefinite integral: $\int\big(\sqrt{1-t^2}\big)^{n-1}\mathrm{d}t$

I would like to evaluate the following indefinite integral $$\int\big(\sqrt{1-t^2}\big)^{n-1}\mathrm{d}t,$$ but -alas- I am not familiarized enough with this kind of integration. I have been suggested to use the substitution $t=\sin\varphi$, but I have some difficulties in applying that. Could you please help me? Thanks a lot!

• Could you be more specific about your difficulties? – Travis Willse Feb 3 '15 at 12:57
• Note too that if $n$ is odd the integrand is just a polynomial. – Travis Willse Feb 3 '15 at 12:58
• @Travis, thanks, I'm looking at this again, I'll come back soon. What about the above comment of user64494? – nullgeppetto Feb 3 '15 at 13:13
• @nullgeppetto :This is the standard notation of the hypergeometric function. – user64494 Feb 3 '15 at 13:28
• @nullgeppetto: Yes, this is a closed form expression. – user64494 Feb 3 '15 at 13:31

By replacing $t$ with $\sin \phi$ we are left with: $$I= \int \cos^n\phi\,d\phi \tag{1}$$ and this integral can be approached by considering the Fourier cosine series of the integrand function: $$\cos^n\phi = \frac{1}{2^n}\left(e^{i\phi}+e^{-i\phi}\right)^n = \frac{1}{2^n}\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{k}2\cos((n-2k)\phi).\tag{2}$$ The RHS of $(2)$ is quite easy to integrate.
• @nullgeppetto: if the integral is considered over $[0,1]$, just replace $t$ with $\sqrt{x}$ to get an integral that depends on the Beta function ($B(a,b)=\frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}$). Otherwise, the integral is given by an incomplete Beta function. – Jack D'Aurizio Feb 4 '15 at 10:57