`int x^2sinx dx` Find the indefinite integral

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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.`


`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)`

Applying`int 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`


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