Is there a general formula for finding the primitive of

$$ \frac{\sin(ax)}{\sin(bx)},$$ where $a,b \in \mathbb R$.

I'm unable to solve this. I tried with integration by parts but that does not work for me. Any hints/ideas?

  • 1
    $\begingroup$ It does not seem that there is a nice general form. See here: wolframalpha.com/input/?i=int+sin%28ax%29%2Fsin%28bx%29+dx $\endgroup$ – Tintarn Sep 12 '15 at 16:29
  • $\begingroup$ It seems it can be done in terms of hypergeometric functions. Of course in some simple cases it will be more explicit. Do you have any feeling that the general case could be expressed in terms of elementary functions only? $\endgroup$ – mickep Sep 12 '15 at 16:31
  • $\begingroup$ @mickep Sorry..I don't.Can you say something more ? $\endgroup$ – Dontknowanything Sep 12 '15 at 16:34
  • $\begingroup$ I used Eulers formula for sine of x and then substituted $\exp{(ix)}$ by u. The integrand turned out to be $-i\cdot u^{b-a-1}\frac{u^{2a}-1}{u^{2b}-1}$. And only in some cases this can be solved in closed form i guess. $\endgroup$ – MrYouMath Sep 12 '15 at 16:50
  • 2
    $\begingroup$ If $b/a$ is rational, and if $c$ is the "greatest common divisor of $a$ and $b$" (so that, in particular, $a = mc$ and $b = nc$ for some integers $m$ and $n$), then $\sin(ax)/\sin(bx)$ is a rational function in $\sin(cx)$ and $\cos(cx)$, and so has an elementary antiderivative by the "tangent half-angle substitution". Otherwise I suspect (without particularly good evidence) the antiderivaticve is not elementary. $\endgroup$ – Andrew D. Hwang Sep 12 '15 at 16:57

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