Answer
$\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi)^{2n+1} (x)^{2n+1}}{(2n+1)!}$
Work Step by Step
Since, we know that the Taylor Series for $\sin x$ is defined as:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...$
Now, $ \sin (\pi x)=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi x)^{2n+1}}{(2n+1)!}=(\pi x)-\dfrac{(\pi x)^3}{3!}+\dfrac{(\pi x)^5}{5!}-\dfrac{(\pi x)^7}{7!}+...$
or, $=\Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi)^{2n+1} (x)^{2n+1}}{(2n+1)!}$
