`int x^3sinx dx` Find the indefinite integral

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marizi eNotes educator| Certified Educator

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`

 

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