Answer
$\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+..$
Work Step by Step
Since, we know that the Maclaurin 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, $\sin x-x+\dfrac{x^3}{3!}=(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+...)-x+\dfrac{x^3}{3!}$
or, $=\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+..$
