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∫ sinx cosx dx .. two results?!! watch

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    I am trying to do ∫ sinx cosx dx
    which I can do, but the confusing thing is I can get two results! :confused:

    if I use substitution
    u = sin x
    du = cos x dx

    then ∫ u du

    which is (u^2)/2
    i.e. (sin^2 x)/2

    but.. what if I used u=cos x?
    du = -sin x dx

    and -∫ u du
    i.e. (-cos^2) /2

    which one is right and why? It can't be both can it? Have I done something wrong?
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    (Original post by kimoni)
    I am trying to do ∫ sinx cosx dx
    which I can do, but the confusing thing is I can get two results! :confused:

    if I use substitution
    u = sin x
    du = cos x dx

    then ∫ u du

    which is (u^2)/2
    i.e. (sin^2 x)/2

    but.. what if I used u=cos x?
    du = -sin x dx

    and -∫ u du
    i.e. (-cos^2) /2

    which one is right and why? It can't be both can it? Have I done something wrong?
    Neither...Do you remember what sin2x is equal to??????2sinxcos
    1/2 sin2x
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    You may just have different C (the constant) values.
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    oh yeah..
    but why can't I do it my way? shouldnt that work too?
    I checked on calc101.com and it said my second answer is right.. what confuses me is why the first one isnt
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    (Original post by kimoni)
    I am trying to do ∫ sinx cosx dx
    which I can do, but the confusing thing is I can get two results! :confused:

    if I use substitution
    u = sin x
    du = cos x dx

    then ∫ u du

    which is (u^2)/2
    i.e. (sin^2 x)/2

    but.. what if I used u=cos x?
    du = -sin x dx

    and -∫ u du
    i.e. (-cos^2) /2

    which one is right and why? It can't be both can it? Have I done something wrong?
    (sin^2x)/2 + A =1/2 -1/2(cos^2x) +A =-(cos^2x/2)+B.
    so they differ by a constant.
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    Integrate them both between limits. You get the same answers. ie they are both right.
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    (Original post by SUKBarracuda)
    Integrate them both between limits. You get the same answers. ie they are both right.
    That's not mathematically too cool.
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    (Original post by fishpaste)
    That's not mathematically too cool.
    in some cases it is (but not in this case)
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    (Original post by SUKBarracuda)
    Integrate them both between limits. You get the same answers. ie they are both right.
    ah ok, thats good to know
 
 
 
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