$A_{mn} = \frac{1}{\pi}\int_0^{\pi}d\theta\, \sin(2m\theta)\, \frac{1-\cos^{2n}(\theta)}{\tan(\theta)} = $ ? $m$, $n$ integers > 0

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 $m=n$, $$A_{mn} = 1- \frac{1}{4^{n}} \sum_{j=1}^{n} \binom{2n+1}{j} = 1 - \frac{1}{4^{n}} \left(4^{n}-1\right) = \frac{1}{4^{n}}. $$

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