Integrate $\int_{0}^{\pi}{-\cos{x}}{_2F_1}\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos{^{2}x}\right)\sin{^{1+n}x}\sin{^{2}x}^{\frac{-1-n}{2}}$

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.

• is there any reason to expect a closed form solution? – tired Nov 28 '15 at 14:19
• functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/… maybe that leads to something – tired Nov 28 '15 at 15:40
• by the way, the sine-terms cancel ,right ? – tired Nov 28 '15 at 15:42
• @tired It's possible but integrate hypergeometric function with the sines and the cosines is rare here. – user294110 Nov 28 '15 at 15:57
• 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 – tired Nov 28 '15 at 16:02

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.

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