Calculus of a Single Variable, Chapter 8, 8.2, Section 8.2, Problem 25

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^3 sin(x) dx , we may apply integration by parts: int u *dv = uv - int v *du .
Let:
u = x^3 then du =3x^2 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^3 sin(x) dx= x^3*(-cos(x)) - int ( -cos(x))* 3x^2dx
= -x^3cos(x)- (-3) int x^2*cos(x) dx
=-x^3cos(x)+3 int x^2 *cos(x) dx
Apply another set of integration by parts on int x^2 *cos(x) dx .
Let:
u = x^2 then du =2x dx
dv= cos(x) dx then v =sin(x)
Note: From the table of integrals, we have int cos(u) du = sin(u) +C .
Applying the formula for integration by parts, we have:
int x^2 cos(x) dx= x^2*(sin(x)) - int sin(x) * (2x) dx
= x^2sin(x)- 2 int x*sin(x) dx
= x^2sin(x)-2 int x *sin(x) dx
Apply another set of integration by parts on int x *sin(x) dx .
Let: u =x then du =dx
dv =sin(x) dx then v =-cos(x)
Note: From the table of integrals, we have int sin(u) du =-cos(u) +C .
int x *sin(x) dx = x*(-cos(x)) -int (-cos(x)) dx
= -xcos(x) + int cos(x) dx
= -xcos(x) + sin(x)
Applying int x *sin(x) dx =-xcos(x) + sin(x) , we get:
int x^2 cos(x) dx=x^2sin(x)-2 int x *sin(x) dx
= x^2sin(x)-2 [-xcos(x) + sin(x)]
=x^2sin(x)+2xcos(x) -2sin(x) .
Applying int x^2 cos(x) dx=x^2sin(x)+2xcos(x) -2sin(x) , we get the complete indefinite integral:
int x^3 sin(x) dx=-x^3cos(x)+3 int x^2 *cos(x) dx
=-x^3cos(x)+3[x^2sin(x)+2xcos(x) -2sin(x)] +C
=-x^3cos(x)+ 3x^2sin(x) +6xcos(x) - 6sin(x) +C