$A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)} = $ ? $m$, $n$ integers > 0 The integral 
$$
A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)}
$$
popped up when I was playing around with the integral representation of the harmonic numbers $H_n$.
I find that Mathematica can give a result for specific values of $m$ and $n$ but I can't get a general formula. Some observations that appear to be true, based on Mathematica results for specific cases:
1) $A_{mn} > 0$.
2) When $m = n$, $A_{nn} = \frac{1}{4^n}$.
3) When $m > n$, $A_{mn} = 0$. (I think this should follow from expanding the $\cos^{2n}(\theta)$ part out in a finite sum of terms like $\sin$ or $\cos$ of integers times $\theta$.)
I tried doing this via contour integration by extending the integral (which is symmetric around $\pi$) to $[0,2\pi)$, and making the substitution $z = e^{i\theta}$. Unless I've made a mistake, this leads to:
$$
A_{mn} = \frac{1}{2\pi i}\oint dz\,
\frac{\left(z^{2m} - z^{-2m}\right)\left(z+z^{-1}\right)\left[1 - \frac{1}{2^{2n}}{\left(z + z^{-1}\right)}^{2n}\right]}{z^2 - 1}\, ,
$$
where the integral is over the counter-clockwise unit circle. This doesn't seem to work: The zeroes appear to cancel on the top and bottom at $z = \pm 1$, so the residues are zero.
Have I made a mistake in the contour integration approach? Is there a better way to evaluate this whole thing?

Using the identities $$\frac{\sin (2m \theta)}{\sin (\theta)} = 2 \sum_{k=0}^{m-1} \cos[(2k+1)\theta]$$
and 
$$\cos^{2n+1}(\theta) = \frac{1}{4^{n}}\sum_{j=0}^{n} \binom{2n+1}{j} \cos [2n+1-2j)\theta], $$ we get 
$$ \begin{align} A_{mn} &= \frac{1}{\pi} \int_{0}^{\pi} \frac{\sin (2 m \theta)}{\sin (\theta)}\left(\cos (\theta) - \cos^{2n+1}(\theta) \right) \, d \theta \\ &= \frac{2}{\pi} \sum_{k=0}^{m-1} \int_{0}^{\pi} \cos[(2k+1)\theta] \cos(\theta) \, d \theta \\ &- \frac{2}{\pi} \, \frac{1}{4^{n}}\sum_{k=0}^{m-1} \sum_{j=0}^{n} \binom{2n+1}{j}\int_{0}^{\pi} \cos[(2k+1)\theta ] \cos[(2n+1-2j) \theta] \, d \theta.  \end{align} $$
First assume that $m <n$.
Then using the fact that 
$$ \int_{0}^{\pi} \cos(mx) \cos(nx) \, dx = \begin{cases} 
      \frac{\pi}{2}  & m = n \\
     0 & \text{otherwise}
   \end{cases} $$ we get
$$A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2}  \right) - \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j} \frac{\pi}{2} = 1 - \frac{1}{4^{n}} \sum_{j=n-m+1}^{n} \binom{2n+1}{j}.  $$
Now if $m> n$, then
$$ A_{mn} = \frac{2}{\pi} \left(\frac{\pi}{2} \right)- \frac{2}{\pi} \, \frac{1}{4^{n}} \sum_{j=0}^{n} \binom{2n+1}{j} \frac{\pi}{2}  = 1 - \frac{1}{4^{n}} \left(4^{n}\right) = 0. \tag{1}$$
And if $m=n$,  $$A_{mn} = 1- \frac{1}{4^{n}} \sum_{j=1}^{n} \binom{2n+1}{j} = 1 - \frac{1}{4^{n}} \left(4^{n}-1\right) = \frac{1}{4^{n}}.  $$

$(1)$ Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$
