`int (x^2+2x)cosx dx`
To evaluate, apply integration by parts `int udv = uv - int vdu` .
So let
`u = x^2+2x`
and
`dv = cosx dx`
Then, differentiate u and integrate dv.
`du = (2x + 2)dx`
and
`v = int cosx dx = sinx`
Plug-in them to the formula of integration by parts. So the integral becomes:
`int (x^2+2x)cosx dx`
`= (x^2+2x)sinx - int sinx * (2x + 2)dx`
`= (x^2 + 2x)sinx...
`int (x^2+2x)cosx dx`
To evaluate, apply integration by parts `int udv = uv - int vdu` .
So let
`u = x^2+2x`
and
`dv = cosx dx`
Then, differentiate u and integrate dv.
`du = (2x + 2)dx`
and
`v = int cosx dx = sinx`
Plug-in them to the formula of integration by parts. So the integral becomes:
`int (x^2+2x)cosx dx`
`= (x^2+2x)sinx - int sinx * (2x + 2)dx`
`= (x^2 + 2x)sinx - int (2x + 2)sinx dx`
To take the integral of (2x + 2sinx)dx, apply integration by parts again.
So let
`u_2 = 2x + 2`
and
`dv_2 = sinx dx`
Differentiate u_2 and integrate dv_2.
`du_2 = 2dx`
and
`v_2 = -cosx`
So the integral becomes:
`= (x^2+2x)sinx - [ (2x + 2)*(-cosx) - int -cosx * 2dx]`
`=(x^2+2x)sinx - [-(2x + 2)cosx + 2int cosx dx]`
`=(x^2+2x)sinx - [-(2x + 2)cosx + 2sinx]`
`= (x^2+2x)sinx +(2x +2)cosx -2sinx `
`= (x^2+2x - 2)sinx + (2x + 2)cosx`
Since the given is indefinite integral, add C.
`= (x^2+2x - 2)sinx + (2x + 2)cosx + C`
Therefore, `int (x^2+2x)cosx dx = (x^2+2x-2)sinx= (x^2+2x - 2)sinx + (2x + 2)cosx + C` .
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