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I would have to integrate $\sin^5(x)\cdot\sin(nx)$, but I have no idea how to. And that's the only coefficient I need for the series.

mrtaurho
  • 16,103
Maaa09
  • 13

2 Answers2

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Write $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$. Then raise to the fifth power and expand using the Binomial Theorem. You will find, using the fact that $(e^{x})^{n}=e^{nx}$ and cancelling appropriate terms, an expression for $\sin^{5}{x}$in terms of trigonometric functions of the form $\sin{nx},\cos{nx}$ ,which is exactly the Fourier expansion of your function.

Noe Blassel
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HINT

Use the half-angle formulae

$$\sin^2\left(\frac x2\right)=\frac{1-\cos(x)}2~~\text{and}~~\cos^2\left(\frac x2\right)=\frac{1+\cos(x)}2$$

Combined with the identities

$$\sin(a)\sin(b)=\frac12(\cos(a-b)-\cos(a+b))~~\text{and}~~\cos(a)\cos(b)=\frac12(\cos(a-b)+\cos(a+b))$$

Start with

$$\int_{-\pi}^\pi\sin^5(x)\sin(nx)\mathrm dx=\int_{-\pi}^\pi\left(\frac{1-\cos(2x)}2\right)^2\sin(x)\sin(nx)\mathrm dx=\dots$$

Can you take it from here?

mrtaurho
  • 16,103
  • Thank you, but it is still in a form that I don't know what to do with. I only learned integration by parts with two functions, this one still has at least four. And I also can't plug it into any calculator, none of them is able to find the fourier series of sin^5(x) or integrate sin^5(x)*sin(nx). – Maaa09 Apr 15 '19 at 16:09
  • @Maaa09 There is no need for Integration By Parts here! As mentioned above you just need to use the half-angle formula to get rid of the powers and afterwards apply the given identities repeatedly. You will end somewhere with basic integration of terms like $\sin(ax)$ and $\cos(bx)$, respectively. It will take some computations, in fact some many computations, but the only part where you integrate will be of the aforemetioned form. – mrtaurho Apr 15 '19 at 17:30
  • Okay, so I think I did that and I managed to divide it to lots of smaller additions and subtractions (I haven't gotten this far before), but in the end every single one of them equaled to 0. – Maaa09 Apr 15 '19 at 18:57
  • @Maaa09 Well done! And it is totally correct that the integral equals zero. WolframAlpha returns $$\int_{-\pi}^\pi\sin^5(x)\sin(nx)\mathrm dx=-\frac{240\sin(n\pi)}{n^6-35n^4+259n^2-225}$$ Which is indeed equal to $0$ as the sine has simple zeros at all integer multplies of $\pi$. To be honest I am not entirely sure what this means in the context of Fourier Series Expansions. – mrtaurho Apr 15 '19 at 19:20
  • I guess it means I am not on the right track. Either way, thank you for your help :) – Maaa09 Apr 15 '19 at 19:37
  • @Maaa09 Glad to help! If you are satisfied with my answer you could also mark it as accepted :) – mrtaurho Apr 15 '19 at 19:37