Recall that $x$ is a fixed element of $]0;1[$. Finally deduce that:
$$\forall y \in ] 0 ; \pi \left[ , \quad \sum _ { n = 1 } ^ { + \infty } \frac { 2 ( - 1 ) ^ { n } y \sin ( y ) } { y ^ { 2 } - n ^ { 2 } \pi ^ { 2 } } = 1 - \frac { \sin ( y ) } { y } . \right.$$