Answer
$\Sigma_{k=0}^{\infty} \dfrac {(-1)^{k}x^{2k+2}}{(2k+1)!} $
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=x \sin x$
Differentiate w.r.t x
$f'(x)=\sin x +x \cos x$
$f''(x)=2 \cos x -x \sin x$
$f'''(x)=-3 \sin x-x \cos x$
$f''''(x)=-4 \cos x +x \sin x$
Now $f(0)=0$, $f'(0)=0$, $f''(0)=2$, $f'''(0)=0$, $f''''(0)=-4$
Hence the Maclaurin Series for the function is
$\displaystyle x^2- \frac {x^4} {6} + .......+ \frac {(-1)^{n}x^{2n+2}}{(2n+1)!} = \Sigma_{k=0}^{\infty} \frac {(-1)^{k}x^{2k+2}}{(2k+1)!} $
