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May I expect the closed-form of this integral?

$$\int_{0}^{\pi}{{_2F_1}\left(\left.\begin{array}{cc}\frac{1}{2}&\frac{-n+1}{2}\\&\frac{3}{2}\end{array}\right|\cos^2(x)\right)(-\cos{x})(\sin{^{n+1}x})(\sin{^{2}x})^{\frac{1}{2}(-n-1)}\,dx}$$

or at least

$$\int{{_2F_1}\left(\left.\begin{array}{cc}\frac{1}{2}&\frac{-n+1}{2}\\&\frac{3}{2}\end{array}\right|\cos^2(x)\right)(-\cos{x})(\sin{^{n+1}x})(\sin{^{2}x})^{\frac{1}{2}(-n-1)}\,dx}$$

The integrand is a result of $$\int{\sin{^{n}(x)}\,dx}.$$

I'm hoping that someone could do the closed-form because my Mathematica couldn't make the result even I tried to put $n=2$ and $n=3$.

Thank you.

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  • $\begingroup$ is there any reason to expect a closed form solution? $\endgroup$ – tired Nov 28 '15 at 14:19
  • $\begingroup$ functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… maybe that leads to something $\endgroup$ – tired Nov 28 '15 at 15:40
  • $\begingroup$ by the way, the sine-terms cancel ,right ? $\endgroup$ – tired Nov 28 '15 at 15:42
  • $\begingroup$ @tired It's possible but integrate hypergeometric function with the sines and the cosines is rare here. $\endgroup$ – user294110 Nov 28 '15 at 15:57
  • $\begingroup$ the point is: as written above the sine terms cancel to $1$. Afterwards u can straightforwardly apply $y=cos(x) $ followed by $y=\sqrt{q}$. the resulting integral seems to be solvable by one of the identities in the link given above $\endgroup$ – tired Nov 28 '15 at 16:02
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May I expect the closed-form of this integral ?

Yes, you may. In fact, the answer is $0$, due to the parity of the sine and cosine functions.


my Mathematica couldn't make the result even when I tried to put $n=2$ and $n=3$.

Mathematica has no problem evaluating the integral, even in its hypergeometric form, once the two sine terms have been reduced.


The integrand is a result of $\displaystyle\int\sin^n(x)~dx.$

If you are already familiar with Mathematica, then you should probably know that it contains a very useful command called FunctionExpand[...] . Applying it to the original integrand yields $\pm~\dfrac12~B\bigg(\cos^2x~,~\dfrac12~,~\dfrac{n+1}2\bigg),~$ where the sign is opposite to that of the cosine function. Again, notice the parity of the integrand. Also, $\displaystyle\int_0^\tfrac\pi2\sin^n(x)~dx$ is a Wallis integral, whose relation to the beta function is well-known.

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  • $\begingroup$ Note: The parity I speak of is taken with regard to $\dfrac\pi2,$ and not with regard to $0,$ since $\cos\bigg(\dfrac\pi2+x\bigg)=-\cos\bigg(\dfrac\pi2-x\bigg).$ $\endgroup$ – Lucian Nov 28 '15 at 21:32
  • $\begingroup$ I don't get what do you do with "FunctionExpand[...]" to prevent the parity of the integrand. I mean how the code in Mathematica will be. $\endgroup$ – user294110 Nov 29 '15 at 0:22
  • $\begingroup$ @user294110: What do you mean by prevent ? $\endgroup$ – Lucian Nov 29 '15 at 0:42
  • $\begingroup$ You wrote "notice the parity of the integrand" so what code I need to input for the "FunctionExpand[...]"? $\endgroup$ – user294110 Nov 29 '15 at 4:31
  • $\begingroup$ @user294110: Your hypergeometric integrand and/or the integral of $\sin^nx$ $\endgroup$ – Lucian Nov 29 '15 at 4:35

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