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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!

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    $\begingroup$ Could you be more specific about your difficulties? $\endgroup$ Feb 3, 2015 at 12:57
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    $\begingroup$ Note too that if $n$ is odd the integrand is just a polynomial. $\endgroup$ Feb 3, 2015 at 12:58
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    $\begingroup$ @Travis, thanks, I'm looking at this again, I'll come back soon. What about the above comment of user64494? $\endgroup$ Feb 3, 2015 at 13:13
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    $\begingroup$ @nullgeppetto :This is the standard notation of the hypergeometric function. $\endgroup$
    – user64494
    Feb 3, 2015 at 13:28
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    $\begingroup$ @nullgeppetto: Yes, this is a closed form expression. $\endgroup$
    – user64494
    Feb 3, 2015 at 13:31

1 Answer 1

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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.

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  • $\begingroup$ @nullgeppetto: See also Wallis' integrals. $\endgroup$
    – Lucian
    Feb 3, 2015 at 15:57
  • $\begingroup$ Hi @Lucian, thanks for your comment! If I'm not mistaken, Wallis integrals are definite, and I cannot find some way to generalize them. Have you something in mind? $\endgroup$ Feb 3, 2015 at 21:22
  • $\begingroup$ Dear Jack, now I can understand your approach. However, I am wondering whether there is something "less complex" than this. I mean that the summation in the Fourier series makes it rather inconvenient for my purposes. Is there any way to express the integral in terms of the Gamma function? Thank you very much! $\endgroup$ Feb 3, 2015 at 21:24
  • $\begingroup$ @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. $\endgroup$ Feb 4, 2015 at 10:57
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    $\begingroup$ Thank you very much @JackD'Aurizio, I'll do as you suggest. I will return after that, so apologies in advance for bothering you again! $\endgroup$ Feb 4, 2015 at 12:09

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