# int x^3sinx 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^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