Closed form for integral $ \int_0^{\pi} \frac{\sin (m \phi)}{(1 + r \cos \phi)^n} d\phi$ Is there a closed form for $n>0$ integer, $m\neq 0$ integer, and $|r|<1$ real?
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\cal I}_{mn}\pars{r} \equiv \int_{0}^{\pi}
     {\sin\pars{m\phi} \over \bracks{1 + r\cos\pars{\phi}}^{n}}\,\dd\phi\,,\qquad
      \mbox{Notice that}\ {\cal I}_{mn}\pars{r} = -{\cal I}_{-m,n}\pars{r}}$

\begin{align}
{\cal I}_{mn}\pars{r} &=
\left.{-\cos\pars{m\phi}/m
       \over
       \bracks{1 +r\cos\pars{\phi}}^{n}}\right\vert_{0}^{\pi}
-
\int_{0}^{\pi}\bracks{-\,{\cos\pars{m\phi} \over m}}\braces{%
-n\,{-r\sin\pars{\phi} \over \bracks{1 + r\cos\pars{\phi}}^{n + 1}}}\,\dd\phi
\\[3mm]&=
{1 \over m}\bracks{%
{\pars{-1}^{m + 1} \over \pars{1 - r}^{n}}
+
{1 \over \pars{1 + r}^{n}}}
+
{nr \over m}\int_{0}^{\pi}
{\cos\pars{m\phi}\sin\pars{\phi}
 \over
 \bracks{1 + r\cos\pars{\phi}}^{n + 1}}\,\dd\phi
\\[3mm]&=
{1 \over m}\bracks{%
{\pars{-1}^{m + 1} \over \pars{1 - r}^{n}}
+
{1 \over \pars{1 + r}^{n}}}
+
{nr \over m}\int_{0}^{\pi}
{\braces{\sin\pars{\bracks{m + 1}\phi} - \sin\pars{\bracks{m - 1}\phi}}/2
 \over
 \bracks{1 + r\cos\pars{\phi}}^{n + 1}}\,\dd\phi
\\[3mm]&=
{1 \over m}\bracks{%
{\pars{-1}^{m + 1} \over \pars{1 - r}^{n}}
+
{1 \over \pars{1 + r}^{n}}}
+
{nr \over 2m}\bracks{{\cal I}_{m + 1,n + 1}\pars{r} - {\cal I}_{m - 1,n + 1}\pars{r}}
\end{align}

$$
{\cal I}_{m + 1,n + 1}\pars{r}
=
-\,{2 \over nr}\bracks{%
{\pars{-1}^{m + 1} \over \pars{1 - r}^{n}} + {1 \over \pars{1 + r}^{n}}}
+
{2m \over nr}\,{\cal I}_{mn}\pars{r} + {\cal I}_{m - 1,n + 1}\pars{r}\,,\qquad
m \geq 1
$$
with
$$
{\cal I}_{0n}\pars{r} = 0 \qquad\mbox{and}\qquad
{\cal I}_{1n}\pars{r}
=
{\pars{1 - r}^{-\pars{n - 1}} - \pars{1 + r}^{-\pars{n - 1}} \over r\pars{n - 1}}\,,
\quad {\cal I}_{-m,n}\pars{r} = -{\cal I}_{mn}\pars{r} 
$$
A: $\int_0^\pi\dfrac{\sin m\phi}{(1+r\cos\phi)^n}d\phi$
$=\int_0^\pi\dfrac{\sin\phi~U_{m-1}(\cos\phi)}{(1+r\cos\phi)^n}d\phi$ (according to http://mathworld.wolfram.com/Multiple-AngleFormulas.html)
$=-\int_0^\pi\dfrac{U_{m-1}(\cos\phi)}{(1+r\cos\phi)^n}d(\cos\phi)$
$=\int_{-1}^1\dfrac{U_{m-1}(x)}{(1+rx)^n}dx$
Can you take it from here?
A: Let $\displaystyle \int_0^{\pi} \sin(m x) \cos^k(x)dx = I(m,k)$. We have
$$I = \int_0^{\pi} \dfrac{\sin(m x)dx}{(1+r \cos(x))^n} = \sum_{k=0}^{\infty} (-r)^k \dbinom{n+k-1}k\int_0^{\pi} \sin(mx) \cos^k(x)dx = \sum_{k=0}^{\infty} (-r)^k \dbinom{n+k-1}k I(m,k)$$
Note that we also have
$$I(m,k) = \int_0^{\pi}\sin(mx) \cos^k(x) dx = \int_0^{\pi}U_{m-1}(\cos(x)) \sin(x) \cos^k(x) dx = \int_{-1}^1 t^k U_{m-1}(t)dt$$
where $U_{m-1}$ is the $(m-1)^{th}$ Chebyshev polynomial of the second kind. Hence, $I(m,k)$ is the $k^{th}$ moment of the $(m-1)^{th}$ Chebyshev polynomial of the second kind. 
