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I'm having some difficulty calculating the integral of $\sin^3\left(\frac{x}{2}\right)\cos^7\left(\frac{x}{3}\right)$ on $[4\pi,16\pi]$. I know the method of odd power in $\sin$ and $\cos$ but this is not the same angle and didn't find a way to get over it. Any hints please?

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  • $\begingroup$ I would guess that the answer is 0 and that you can prove this using a symmetry argument or change of variables without actually having to find the antiderivative (which looks hard). $\endgroup$ – Stefan Smith Nov 3 '13 at 18:28
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By substituting $x=6t$, we get $$ \int_{4\pi}^{16\pi}\sin^3\left(\frac{x}2\right)\cos^7\left(\frac{x}3\right)\,dx=6\int_{2\pi/3}^{2\pi+2\pi/3}\sin^33t\,\cos^72t\,dt\\=6\int_{0}^{2\pi}\sin^33t\,\cos^72t\,dt=6\int_{-\pi}^{\pi}\sin^33t\,\cos^72t\,dt $$ (note that $\sin^33t\,\cos^72t$ has period $2\pi$). Now we see that the integral is zero because the integrand is odd.

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  • $\begingroup$ Are you sure about this? $\endgroup$ – Ron Gordon Nov 3 '13 at 19:20
  • $\begingroup$ Yes.${\ \ \ \ \ }$ $\endgroup$ – Martin Argerami Nov 3 '13 at 20:27
  • $\begingroup$ I just don't see it. The frequencies of the sine and cosine are different in the problem statement. $\endgroup$ – Ron Gordon Nov 3 '13 at 20:31
  • $\begingroup$ they are (almost) multiples of the same frequency $\endgroup$ – Thomas Nov 3 '13 at 20:34
  • $\begingroup$ Now I see what you mean, sorry. I had several typos, they should be corrected now. $\endgroup$ – Martin Argerami Nov 3 '13 at 20:35
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Hint: Try to make one of the inside argument equal to the other and use the angle addition or subtraction formula to expand.

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Here is an approach.

i) Use the identity

$$ e^{it}=\cos t+i\sin t .$$

ii) use the binomial theorem

$$ (a+b)^n = \sum_{k=0}^{n} {n\choose k} a^k b^{n-k}. $$

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  • $\begingroup$ This leads nowhere $\endgroup$ – Thomas Nov 3 '13 at 20:34
  • $\begingroup$ @Mhenni: I would like to see how you use this approach to calculate the integral. $\endgroup$ – Martin Argerami Nov 5 '13 at 17:24

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