Answer
$-\frac{1}{2}\cos{x^2}+C$
Work Step by Step
$\int x\sin{x^2}dx$
Let $u=x^2\hspace{8mm}du=2xdx\hspace{8mm}dx=\frac{du}{2x}$
Therefore,
$\int x\sin{x^2}dx=\int x\sin{u}\frac{du}{2x}=\frac{1}{2}\int\sin{u}du$
$=\frac{1}{2}(-\cos{u})+C=-\frac{1}{2}\cos{x^2}+C$
