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    (Original post by Bigcnee)
    pretty pointless.
    it was just a suggestion, i haven't done it myself
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    (Original post by king of swords)
    eeeek....i got (by parts):
    Code:
    (2/3)(xsinx + (1 - x)cosx)e^x) + c
    :confused:
    Check it by differentiating, there may be different ways of writing the same answer.

    I differentiated my answer, and it turned out right.
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    (Original post by mikesgt2)
    Check it by differentiating, there may be different ways of writing the same answer.

    I differentiated my answer, and it turned out right.
    oh yeah!....i wonder how i got the factor 2/3 in there then :confused:
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    The following is a smart-arse solution.

    Everyone knows that

    (int) E^(u*x)*Sin[x] dx
    = (E^(u*x)*(-Cos[x] + u*Sin[x])) / (1 + u^2) + constant.

    Differentiating,

    (d/du) [(int) E^(u*x) Sin[x] dx]
    = [E^(u*x)*(-((-2*u + x + u^2*x)*Cos[x])
    + (1 - u^2 + u*x + u^3*x)*Sin[x])] / (1 + u^2)^2.

    We're allowed to do (d/du) and (int) in either order, so the left-hand side equals

    (int) x*E^(u*x)*Sin[x] dx + constant

    Set u = 1 on both sides to get

    (int) x*E^x*Sin(x) dx
    = (E^x*(Cos[x] - x*Cos[x] + x*Sin[x])) / 2 + constant

    Jonny W.
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    im a little confused - your differentiating the integral (I) with respect to u, but I is defined implictily as f(x)g(u). Yet when you do dI/du you should have some dx/du terms, shouldnt you? Plus im not exactly sure how you can just say LHS = [int] xI dx with little justification. You are probably right - you have the right answer but im not sure why...?
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    (Original post by It'sPhil...)
    im a little confused - your differentiating the integral (I) with respect to u, but I is defined implictily as f(x)g(u). Yet when you do dI/du you should have some dx/du terms, shouldnt you? Plus im not exactly sure how you can just say LHS = [int] xI dx with little justification. You are probably right - you have the right answer but im not sure why...?
    Define a function I by

    I(u, x) = (int from 0 to x) E^(u*y)*Sin[y] dy

    so that, integrating by parts twice,

    I(u, x) = (1 - E^(u*x)*Cos[x] + E^(u*x)*u*Sin[x])/(1 + u^2).

    Then, using basic calculus,

    (d/du) I(u, x)
    = (-2*u - E^(u*x)*(-2*u + x + u^2*x)*Cos[x]
    + E^(u*x)*(1 - u^2 + u*x + u^3*x)*Sin[x])/(1 + u^2)^2.

    (We're differentiating wrt u, keeping x constant.) Also, swapping (d/du) and the integral sign,

    (d/du) I(u, x)
    = (int from 0 to x) (d/du) E^(u*y)*Sin[y] dy
    = (int from 0 to x) y*E^(u*y)*Sin[y] dy.

    So, for all u and x,

    (int from 0 to x) y*E^(u*y)*Sin[y] dy
    = (-2*u - E^(u*x)*(-2*u + x + u^2*x)*Cos[x]
    + E^(u*x)*(1 - u^2 + u*x + u^3*x)*Sin[x])/(1 + u^2)^2.

    Setting u = 1 gives, for all x,

    (int from 0 to x) y*E^(y)*Sin[y] dy
    = (-1 - E^x*(-1 + x)*Cos[x] + E^x*x*Sin[x])/2.

    That's the same answer as before, give or take a constant.

    -

    The step in which I swapped (d/du) and the integral sign needs justification to be rigorous. But you don't need to know the justification to believe the answer. You need only differentiate

    (-1 - E^x*(-1 + x)*Cos[x] + E^x*x*Sin[x])/2

    and check that you get

    x*E^(x)*Sin[x].

    Jonny W.
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    (Original post by byb3)
    Hi,

    This is question 6 on my exercise sheet which you would assume to be easier than the other 10 questions after it. However I can't do it, I get repetitive motion when doing it in parts with substitution. What sort of method works?

    Integrate(2*x*sin(x)*exp(x)) with respect to x of course.

    Good luck to anyone who dares. I personally think that the question might be easier than it appears to be, just I am using the wrong method.

    Adam
    Wohaaa people, your doing it way to complicated.

    First of all, lets get the 2 outside the integral:

    2 integral sin x e^x dx

    Nos wed o it by parts and we get:

    2 ( sin x * e^x - integral -cos x e^x dx ) =

    2 ( sin (x) e^x + integral cos x e^x dx) | use by parts again:

    2 ( sin(x) * ( e^x + cos x - e^x integral sin x e^x dx ))

    Resolve the paranthesis and we get :

    2 sin (x) * 2 cos (x ) - 2 integral sin x e^x dx

    Now we have the equation

    2 integral sin x e^x dx = 2 ( sin x cos x) - 2 integral sin x e^x dx

    So we may just add "2 integral sin x e^x dx" on both sides to get

    4 integral sin x e^x dx = 2 (sin x cos x )

    Now divide by two on both sides and we are left with

    integral 2 sin x e^x dx = sin x cos x

    And of course we must not forget the constant:

    integral 2 sin x e^x dx = sin x cos x + C
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    (Original post by Jonatan)
    Wohaaa people, your doing it way to complicated.

    First of all, lets get the 2 outside the integral:

    2 integral sin x e^x dx

    Nos wed o it by parts and we get:

    2 ( sin x * e^x - integral -cos x e^x dx ) =

    2 ( sin (x) e^x + integral cos x e^x dx) | use by parts again:

    2 ( sin(x) * ( e^x + cos x - e^x integral sin x e^x dx ))

    Resolve the paranthesis and we get :

    2 sin (x) * 2 cos (x ) - 2 integral sin x e^x dx

    Now we have the equation

    2 integral sin x e^x dx = 2 ( sin x cos x) - 2 integral sin x e^x dx

    So we may just add "2 integral sin x e^x dx" on both sides to get

    4 integral sin x e^x dx = 2 (sin x cos x )

    Now divide by two on both sides and we are left with

    integral 2 sin x e^x dx = sin x cos x

    And of course we must not forget the constant:

    integral 2 sin x e^x dx = sin x cos x + C
    I'm afraid you've done the wrong integral. You've missed a factor of x from the original question.
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    im on my holiday.... i'm not meant to be thinking! (otherwise i would have helped out)
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    (Original post by rahaydenuk)
    I'm afraid you've done the wrong integral. You've missed a factor of x from the original question.
    Bah! This reminds me on what happened when russel discovered his famous barberer paradox...
 
 
 
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