Tricky Sine integral

1. Jul 7, 2013

klawlor419

1. The problem statement, all variables and given/known data

Any ideas for how to solve the following integral?

$$\int_{0}^{\pi}\sin{n x}\sin{x}^3 dx$$

where n is a positive integer

2. Relevant equations

Various sine and cosine identities

3. The attempt at a solution

I haven't much of a clue how to solve the integral. Its an odd function times an odd function which gives an even function, over a symmetric range (at least symmetric for the Sine function or perhaps portions of the function's n-values).

I tried clearing out a sin^2 from the integral by using the double-angle formula. It didn't really break down into anything that lead to any obvious results for the integrals.

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2013

vanhees71

I guess that this is a typo, because I don't think that you can find a closed solution for that integral. So I think you want to solve
$$I=\int_0^{\pi} \mathrm{d} x \; \sin(n x) (\sin x)^3.$$
In this case it may be helpful to use
$$(\sin x)^3=\frac{1}{4}[3\sin x - \sin(3 x)].$$

3. Jul 7, 2013

klawlor419

Yes that was a typo, sorry about that.

Cheers, I think I have done something like that already. I used the exponential forms for the sine functions and factored them all together. I am pretty sure that this gives that identity. In fact, it is very similar.

Once I factored together the terms, I found that only for the values of n=1 and n=3 does the integral have any value at all. Which are precisely the coefficients inside of the identity.

Though I am not completely satisfied with this answer. Because now it requires going back to the original and solving the integral for those specific values.