*Replied by ex-member ***aziiri**:Grigorios Kostakos wrote:

A possible first step is to use Lagrange's trigonometric identity: \begin{align*}

\mathop{\sum}\limits_{k=1}^n {\sin(kx)}&=\frac{1}{2}\frac{\cos\frac{x}{2}-\cos\bigl({\bigl({n +\frac{1}{2}}\bigr) x}\bigr)}{\sin\frac{x}{2}}\\

&=........................\\

&=\frac{\sin\bigl({({n+1})\frac{x}{2}}\bigr)\,\sin\bigl({n\frac{x}{2}}\bigr)}{\sin\frac{x}{2}}\,.

\end{align*}

It is easier that that, note that for two integers \(n\neq m \) : \[\int_{-\pi}^{\pi} \sin n x\sin m x \ \mathrm{d}x = \frac{1}{2} \int_{-\pi}^{\pi} \cos((n-m) x) +\cos ((n+m)x) \ \mathrm{d}x = 0\]

Then : \[\begin{align*}\int_{-\pi}^{\pi } \left(\sum_{k=1}^{2014} \sin k x\right)^2 \ \mathrm{d}x &=\int_{-\pi}^{\pi} \left(\sum_{k=1}^{2014} \sin^2 kx\right)+2 \left(\sum_{1\leq i<j\leq 2014} \sin ix \sin jx\right) \ \mathrm{d}x \\ &=\sum_{k=1}^{2014} \int_{-\pi}^{\pi} \sin^2 kx \ \mathrm{d}x \\ &= \frac{1}{2} =\sum_{k=1}^{2014} \int_{-\pi}^{\pi} 1-\cos kx \ \mathrm{d}x \\ &= 2014\pi.\end{align*}\]