int x^2sinx dx Find the indefinite integral

Recall that indefinite integral follows int f(x) dx = F(x) +C where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
 For the given  integral problem: int x^2 sin(x) dx , we may apply integration by parts: int u *dv = uv - int v *du.
Let:
u = x^2  then du =2x dx  
dv= sin(x) dx then v = -cos(x)
Note: From the table of integrals, we have int sin(u) du = -cos(u) +C .
Applying the formula for integration by parts, we have:
int x^2 sin(x) dx= x^2*(-cos(x)) - int ( -cos(x))* 2x dx
                              = -x^2cos(x)- (-2) int x*cos(x) dx
                              =-x^2cos(x)+2 int x *cos(x) dx
Apply another set of integration by parts on int x *cos(x) dx .
Let: u =x then du =dx
       dv =cos(x) dx then v =sin(x)
Note: From the table of integrals, we have int cos(u) du =sin(u) +C .
int x *cos(x) dx = x*sin(x) -int sin(x) dx
                              = xsin(x) -(-cos(x))  
                              = xsin(x) + cos(x)
Applyingint x *cos(x) dx =xsin(x) + cos(x) , we get the complete indefinite integral as:
int x^2 sin(x) dx=-x^2cos(x)+2 int x *cos(x) dx
                              =-x^2cos(x)+2 [xsin(x) + cos(x)]+C
                               =-x^2cos(x)+2xsin(x) +2cos(x) +C