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

Question

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.

Answer 1

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.

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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.

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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.