Answer
$x^2-\dfrac{x^3}{2}+\dfrac{x^4}{6}-....$
Work Step by Step
The Maclaurin series for $\sin x$ can be defined as:
$x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+......=\Sigma_{n=0}^{\infty} \dfrac{ (-1)^n x^{2n+1}}{(2n+1)!}$
Therefore,
$x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-......=\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{(n+1)}$
This implies that $f(x)=\sin x \ln (1+x)$
$\sin x \times \ln (1+x)=[\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}]-[\Sigma_{n=0}^{\infty} (-1)^n \dfrac{x^{n+1}}{(n+1)}]$
and $\sin x \times \ln (1+x)=x^2-\dfrac{x^3}{2}+\dfrac{x^4}{6}-....$
