Fallacious moving of powers resulting with a correct trigonometric series identity. 
Prove that
  $$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} 
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}
$$
  for all values of $x$.

I came across this joke which said that you could just bring the powers out of the brackets and everything works out right.
$$
\require{cancel}
\sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} 
( \sin^{2(\textbf{n+r+1})}x + \cos^{2(\textbf{n+r+1})}x )\right)
\\ = \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} 
(\cancelto1{\sin^2x + \cos^2x })^{\textbf{n+r+1}}\right)
\\ = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1}
$$
But how would you actually go about proving this? I noticed that this value is also equal to $B(n+1,n+1)$, which may or may not be relevant.

Let's prove $$
\\ \sum_{r=0}^n \left( \frac{ (-1)^r {n \choose r} } {n+r+1} 
( \sin^{2(n+r+1)}x + \cos^{2(n+r+1)}x )\right) = \sum_{r=0}^n \frac{ (-1)^r {n \choose r} } {n+r+1} \tag1
$$
for all values of $x$.
By differentiating both sides with respect to $x$, on the right hand side one gets $0$, on the left hand side one gets
$$
\sum_{r=0}^n (-1)^r {n \choose r} 
\left( 2\cos x\cdot\sin^{2(n+r+1)-1}x - 2\sin x\cdot\cos^{2(n+r+1)-1}x \right)
$$
$$
2\cos x\cdot\sin^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} 
\sin^{2r}x - 2\sin x\cdot\cos^{2n+1}x \sum_{r=0}^n (-1)^r {n \choose r} 
\cos^{2r}x
$$ $$
2\cos x\cdot\sin^{2n+1}x\cdot \left(1-\sin^2x \right)^n-2\sin x\cdot\cos^{2n+1}x\cdot \left(1-\cos^2x \right)^n=0.
$$
Since both sides of $(1)$ clearly agree at $x=0$, then $(1)$ is true for all values of $x$.
