Sunday, January 24, 2016

`int (x^2 + 2x) cos(x) dx` Evaluate the integral

`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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