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Integrand

Given that $\displaystyle{ \sum_{k=1}^{n}k = \frac{\frac{n}{2}\cdot \frac{n+1}{2}}{\frac{1}{2}}}$, we have $\displaystyle{ \sum_{k=1}^{n} \sin(k) = \frac{\sin(\frac{n}{2})\cdot\sin(\frac{n+1}{2})}{\sin(\frac{1}{2})} }$.

Proof: take the sine of everything.

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