Answer
$(x^2-2) \cos (1-x) +2x \sin (1-x) +C$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int x^2 \sin (1-x) dx=x^2 \cos (1-x) -\int 2x \cos (1-x) dx$...(1)
Consider $A=\int 2x \cos (1-x) dx$
Apply Integration by parts formula which suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ on part A.
Thus, we have
$\int 2x \cos (1-x) dx=2x \sin (1-x) -\int 2 \sin (1-x) dx=2x \sin (1-x) -\ 2 \cos (1-x) +c'$
Now, from equation (1), we get
$x^2 \cos (1-x)+2x \sin (1-x) -\ 2 \cos (1-x) +c'=(x^2-2) \cos (1-x) +2x \sin (1-x) +C$
