# Find the Fourier series of $\sin^3(x)$ on $[-\pi,\pi]$

I'm having trouble integrating $B_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}\sin^{3}t \,\sin(nt)\,dt$.

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I suggest using trig identities to rewrite $\sin^3(x)$ as $A\sin(x)+B\sin(2x)+C\sin(3x)$. Then there's your series right there.

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Hint: Use: 1) $4\sin^3(t)=3 \sin(t)-\sin(3t)$

2) Use the formula for $\sin(a)\sin(b)$

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After using that identity, I get that $B_n=0$. Isn't this suppose to be non-zero since $sin^3(\theta)$ is an odd function? –  user80979 Apr 14 '14 at 21:49
There is nothing more to compute, the Fourier series is nothing more nor less than that sum of sine terms (divided by 4). –  LutzL Apr 14 '14 at 21:56
@LutzL, But then $A_0=0$, so is $B_n$, then how do we represent the function as a trigonometric polynomial? –  user80979 Apr 14 '14 at 22:21
I see $B_1=\frac34$ and $B_3=-\frac14$, your function is already represented as trigonometric polynomial? –  LutzL Apr 14 '14 at 23:07
@LutzL, Don't we get 0/0 for $n=1,3$? –  user80979 Apr 15 '14 at 20:02