May I expect the closed-form of this integral?


or at least


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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.