Answer
$\Sigma_{k=0}^{\infty} \dfrac {(-1)^k\pi^{2k+1} x^{2k+1}} {(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!} + ..\\
$
$f(x)=\sin \pi x$
$f'(x)=\pi cos\pi x$
$f''(x)=-\pi^2 \sin \pi x$
$f'''(x)=-\pi^3 \cos \pi x$
$f''''(x)=\pi^4 \sin \pi x$
Now $f(0)=\pi$ , $f'(0)=\pi$ , $f''(0)=0$, $f'''(0)=-\pi^3 $, $f''''(0)=0$
Hence the Maclaurin Series for the function is
$\displaystyle \pi x - \frac {\pi^3 x^3} {2!} + .......+ \frac {(-1)^3\pi^{2n+1} x^{2n+1}} {(2k+1)!} = \Sigma_{k=0}^{\infty} \frac {(-1)^k\pi^{2k+1} x^{2k+1}} {(2k+1)!} $
