If $\int { x\sin { x } ; } dx=-x\cos { x } +\alpha$, then $\alpha$ is equal to

$I=\displaystyle\int{x\sin{x}dx}$

Integrating by parts,

Let $u=x\Rightarrow\,du=dx$

$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$......by parts formula

$I=-x\cos{x}+\displaystyle\int{\cos{x}dx}$

$I=-x\cos{x}+\sin{x}+c$ ......where $c$ is the constant of integration.

We have $I=\displaystyle\int{x\sin{x}dx}=-x\cos{x}+\alpha$

Comparing with $I=-x\cos{x}+\sin{x}+c$ we have

$\alpha=\sin{x}+c$