Integral of $\sin^3\left(\frac{x}{2}\right)\cos^7\left(\frac{x}{3}\right)$

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?

• 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). – Stefan Smith Nov 3 '13 at 18:28

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.

• Yes.${\ \ \ \ \ }$ – Martin Argerami Nov 3 '13 at 20:27
• I just don't see it. The frequencies of the sine and cosine are different in the problem statement. – Ron Gordon Nov 3 '13 at 20:31
• they are (almost) multiples of the same frequency – Thomas Nov 3 '13 at 20:34
• Now I see what you mean, sorry. I had several typos, they should be corrected now. – Martin Argerami Nov 3 '13 at 20:35

Hint: Try to make one of the inside argument equal to the other and use the angle addition or subtraction formula to expand.

Here is an approach.

i) Use the identity

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

ii) use the binomial theorem

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

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