Evaluating $\int_{0}^{\pi}\frac{\cos^m(x/2)}{1+\sin(x/2)} \frac{\sin^n(x/2)}{\ln\sin(x/2)}dx$ We wish to evaluate this integral,
$$\int_{0}^{\pi}\frac{\cos^m(x/2)}{1+\sin(x/2)}\cdot \frac{\sin^n(x/2)}{\ln\sin(x/2)}\mathrm dx=F(m,n)$$
We also know the closed form for:
$F(3,n)=-2\ln\left(\frac{n+2}{n+1}\right)$
$F(5,n)=-2\ln\left(\frac{(n+2)(n+3)}{(n+1)(n+4)}\right)$
This is not a contest integral, it is just made up integral.
 A: Note this solution is only for odd $m>1$. Using the substitution $u=\frac{x}{2}$ followed by $t=\sin(u)$
$$I(m,n)=\int_{0}^\pi \frac{\cos^m(x/2)\sin^n(x/2)}{(1+\sin(x/2))\ln(\sin(x/2))}dx=$$ $$2\int_{0}^1\frac{(1+t)^{\frac{m-1}{2}-1}(1-t)^{\frac{m-1}{2}}t^n}{\ln(t)}dt$$
Now using the binomial theorem, $$I_n(m,n)=2\int_{0}^1(1+t)^{\frac{m-1}{2}-1}(1-t)^{\frac{m-1}{2}}t^ndx=$$
$$2\sum_{v=0}^{\frac{m-3}{2}}\sum_{w=0}^{\frac{m-1}{2}}\binom{\frac{m-3}{2}}{v}\binom{\frac{m-1}{2}}{w}(-1)^w\int_{0}^1 t^{v+w+n}=$$
$$2\sum_{v=0}^{\frac{m-3}{2}}\sum_{w=0}^{\frac{m-1}{2}}\binom{\frac{m-3}{2}}{v}\binom{\frac{m-1}{2}}{w}(-1)^w\frac{1}{1+v+w+n}$$
This implies,
$$I(m,n)=2\sum_{v=0}^{\frac{m-3}{2}}\sum_{w=0}^{\frac{m-1}{2}}\binom{\frac{m-3}{2}}{v}\binom{\frac{m-1}{2}}{w}(-1)^w\ln(1+v+w+n)+f(m)$$
For some function $f(m)$. I'm a bit tired, so I'm going to stop here. Numerically it seems that this agrees for values up to about $1<m<23$ if $f(m)=0$, but then it diverges slowly off. I'm not entirely sure if it's my calculator or if $f(m)$ is causing it to diverge. Also, the above sum can most likely be simplified to what I wrote in the comments. Another note is that I'm not really sure why this applies to odd integers greater than $1$. It could have to do with $f(m)$.
