HOW TO INTEGRATE THIS>>> (x^2) (e^3x) sin(3x) dx?

Question

integrate (x^2) (e^3x) sin(3x) dx

i tried integration by parts... but i think it wasn't the right thing to do...
any idea how to do this?

Answer 1

u.............v'

x²............e^(3x)sin3x

2x...........1/6 e^(3x)(sin3x-cos3x)

2............. - 1/18 e^(3x)cos3x

0............. -1/108 e^(3x)(sin3x+cos3x)



Thus, the integral

=1/6 x²e^(3x)(sin3x-cos3x) +1/9 xe^(3x)cos3x - 1/54 e^(3x)(sin3x+cos3x) + C



You can rewrite the answer to other forms but I will just leave it like that.



With this method, I assume you know how to integrate e^(3x)sin3x or e^(3x)cos3x

You just need to do integration by parts, letting u = e^3x, v' = sin3x or cos3x

Answer 2

The answer is (1/54)e^(3x)[(9x^2-1)sin(3x) - (1-3x)^2cos(3x)]+C.

It is not easy to do by parts, but you could try integrating (x^2)(e^(3x)) by parts

and if this is f(x) , integrate f(x)sin(3x) by parts.

Answer 3

First, by Chain Rule, we have to determine the following derivatives

d/dx (x² sinx) = x² cosx + 2x sinx

d/dx (x² cosx) = -x² sinx + 2x cosx

d/dx (x cosx) = -x sinx + cosx



Simplify the expression



∫ x² e^(3x) sin(3x) dx, let y = 3x

= (1/27) ∫ y² siny e^y dy



The trick for each integration by parts is to isolate the exponential term as the integrating factor



∫ y² siny e^y dy, Integrate by Parts, u = y² siny, dv = e^y dy

= ∫ u dv

= uv - ∫ v du

= y² siny e^y - ∫ y² cosy e^y dy - 2 ∫ y siny e^y dy ............... [1]



∫ y² cosy e^y dy, Integrate by Parts, u' = y² cosy, dv' = e^y dy

= u' v' - ∫ v' du'

= y² cosy e^y + ∫ y² siny dy - 2 ∫ y cosy e^y dy ................. [2]



Substitute [2] into [1]:

∫ y² siny e^y dy = y² siny e^y - [ y² cosy e^y + ∫ y² siny dy - 2 ∫ y cosy e^y dy ] - 2 ∫ y siny e^y dy

.

∫ y² siny e^y dy = y² siny e^y - y² cosy e^y - ∫ y² siny dy + 2 ∫ y cosy e^y dy - 2 ∫ y siny e^y dy

.

2 ∫ y² siny e^y dy = y² siny e^y - y² cosy e^y + 2 ∫ y cosy e^y dy - 2 ∫ y siny e^y dy

.

∫ y² siny e^y dy = (1/2) y² siny e^y - (1/2) y² cosy e^y + ∫ y cosy e^y dy - ∫ y siny e^y dy ........... [3]



∫ y cosy e^y dy, Integrate by Parts, let U = y cosy, dV = e^y dy

= ∫ U dV

= UV - ∫ V dU

= y cosy e^y + ∫ y siny e^y dy - ∫ cosy e^y dy .................. [4]



Substitute [4] into [3]:

∫ y² siny e^y dy

= (1/2) y² siny e^y - (1/2) y² cosy e^y + y cosy e^y + ∫ y siny e^y dy - ∫ cosy e^y dy - ∫ y siny e^y dy

= (1/2) y² siny e^y - (1/2) y² cosy e^y + y cosy e^y - ∫ cosy e^y dy ............... [5]



∫ cosy e^y dy, Integrate by Parts, U' = cosy, dV' = e^y dy

= ∫ U' dV'

= U' V' - ∫ V' dU'

= cosy e^y + ∫ siny e^y dy, Integrate by Parts, U'' = siny, dV'' = e^y dy

= cosy e^y + ∫ U'' dV''

= cosy e^y + U'' V'' - ∫ V'' dU''

= cosy e^y + siny e^y - ∫ cosy e^y dy



==> 2 ∫ cosy e^y dy = cosy e^y + siny e^y

==> ∫ cosy e^y dy = (1/2) cosy e^y + (1/2) siny e^y ................... [6]



Substitute [6] into [5]:

∫ y² siny e^y dy

= (1/2) y² siny e^y - (1/2) y² cosy e^y + y cosy e^y - (1/2) cosy e^y - (1/2) siny e^y

= (1/2) y² siny e^y - (1/2) y² cosy e^y + (1/2) cosy e^y - (1/2) siny e^y



Substitute by y = 3x and add an arbitrary constant C and you're done

∫ x² e^(3x) sin(3x) dx

= (1/27) ∫ y² siny e^y dy

= (1/54) [ y² siny e^y - y² cosy e^y + cosy e^y - siny e^y ] + C

= (1/54) [ 9x² sin(3x) e^(3x) - 9x² cos(3x) e^(3x) + cos(3x) e^(3x) - sin(3x) e^(3x) ] + C

= (1/54) e^(3x) [ 9x² sin(3x) - 9x² cos(3x) + cos(3x) - sin(3x) ] + C

= (1/54) e^(3x) [ (9x² - 1) sin(3x) - (9x² - 1) cos(3x) ] + C

= (√2 /54) (9x² - 1) sin(3x - π/4) e^(3x) + C

Answer 4

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Answer 5

cos θ = {(e^iθ) + (e^-iθ)} / 2 (where i = √ -1)



i sin θ = {(e^i θ) - (e^-i θ)} / 2



=> sinθ = {(e^i θ) - (e^-i θ)} / 2i



=> sin3x = {(e^3ix) - (e^-3ix)} / 2i



=> (e^3x) sin(3x) = (e^3x) X {(e^3ix) - (e^-3ix)} / 2i



=> (e^3x) sin(3x) = [{e^3x(i+1)} - {e^3x(1-i)}] / 2i



=> (x^2) (e^3x) sin(3x) = [x^2{e^3x(i+1)} - x^2{e^3x(1-i)}] / 2i



∫(x^2) (e^3x) sin(3x) dx =(1 / 2i) ∫ x^2{e^3x(i+1)}dx - (1 / 2i) ∫ x^2{e^3x(1-i)}dx



3(i+1) and 3(i-1) are constants