Answer
$c_{2}=1,c_{1}=1$
Work Step by Step
Since $F$ is an antiderivative of $f$ it follows:
$$F'(x)=f(x)$$
$$(c_{1}x\sin(x)+c_{2}\cos(x))'=x\cos{(x)}$$
$$(c_{1}x\sin(x))'+(c_{2}\cos(x))'=x\cos{(x)}$$
$$c_{1}(x\sin(x))'+c_{2}(\cos(x))'=x\cos{(x)}$$
$$c_{1}(x)'\sin(x)+c_{1}x(\sin(x))'+c_{2}(\cos(x))'=x\cos{(x)}$$
$$c_{1}\cdot 1\cdot \sin(x)+c_{1}x\cos(x)-c_{2}\sin(x)=x\cos{(x)}$$
$$c_{1}\cdot \sin(x)+c_{1}x\cos(x)-c_{2}\sin(x)=x\cos{(x)}$$
$$(c_{1}-c_{2})\cdot \sin(x)+c_{1}x\cos(x)=x\cos{(x)}$$
$$(c_{1}-c_{2})\cdot \sin(x)+c_{1}x\cos(x)=0\cdot\sin(x)+ x\cos{(x)}$$
By identification of factors of both sides it follows:
$$c_{1}-c_{2}=0,c_{1}=1$$
$$1=c_{2},c_{1}=1$$
