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    Hello, this question of mine is related to integration unit of C4:

    Integration using substitution,
    When we end up with u as the final thing , and the substitution we made initially was x=tan u , how do we get the answer back in terms of x ?

    because we surely cannot leave the final answer in forms of u right?
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    (Original post by laurawoods)
    Hello, this question of mine is related to integration unit of C4:

    Integration using substitution,
    When we end up with u as the final thing , and the substitution we made initially was x=tan u , how do we get the answer back in terms of x ?

    because we surely cannot leave the final answer in forms of u right?
    For definite integrals yes. If indefinite then no, you must convert back using arctan.
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    (Original post by laurawoods)
    Hello, this question of mine is related to integration unit of C4:

    Integration using substitution,
    When we end up with u as the final thing , and the substitution we made initially was x=tan u , how do we get the answer back in terms of x ?

    because we surely cannot leave the final answer in forms of u right?
    u=\tan^{-1}x
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    (Original post by Mr M)
    u=\tan^{-1}x
    thanks to both answers- joostan and Mr M. !
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    (Original post by laurawoods)
    thanks to both answers- joostan and Mr M. !
    np
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    (Original post by joostan)
    np
    Pls can u answer this maths question of mine

    Hello, if we have something like y+1 to integrate, don't we get y^2/2 + y


    Pls can you look at the solution to question no 4 on this paper:
    http://papers.xtremepapers.com/Edexc...d%20H%20MS.pdf

    They went about a different way ...do u think my method would be fine ...because I got the same answer as them >...
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    (Original post by laurawoods)
    Pls can u answer this maths question of mine

    Hello, if we have something like y+1 to integrate, don't we get y^2/2 + y


    Pls can you look at the solution to question no 4 on this paper:
    http://papers.xtremepapers.com/Edexc...d%20H%20MS.pdf

    They went about a different way ...do u think my method would be fine ...because I got the same answer as them >...
    If you get the same answer then I imagine its fine.
    You'll have to ask MrM
    (Original post by Mr M)
    ...
    to be 100% sure but I think they should have put:
    \int y + 1 \ d(y+1) = - \int x -2 \ d(x-2)
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    (Original post by Mr M)
    u=\tan^{-1}x
    Hello Mr M , Joostan asked me to ask you about this:
    If we have something like y+1 to integrate, don't we get y^2/2 + y


    Pls can you look at the solution to question no 4 on this paper:
    http://papers.xtremepapers.com/Edexc...d%20H%20MS.pdf

    They went about a different way ...do u think my method would be fine ...because I got the same answer as them >...[/B]
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    (Original post by laurawoods)
    Pls can u answer this maths question of mine

    Hello, if we have something like y+1 to integrate, don't we get y^2/2 + y


    Pls can you look at the solution to question no 4 on this paper:
    http://papers.xtremepapers.com/Edexc...d%20H%20MS.pdf

    They went about a different way ...do u think my method would be fine ...because I got the same answer as them >...
    both methods are fine - they've just used "recognition" because d/dy(y+1)^2 = 2(y+1)
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    (Original post by laurawoods)
    Hello Mr M , Joostan asked me to ask you about this:
    If we have something like y+1 to integrate, don't we get y^2/2 + y


    Pls can you look at the solution to question no 4 on this paper:
    http://papers.xtremepapers.com/Edexc...d%20H%20MS.pdf

    They went about a different way ...do u think my method would be fine ...because I got the same answer as them >...[/B]
    Your answer is fine (if you remembered the constant of integration).
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    (Original post by Mr M)
    Your answer is fine (if you remembered the constant of integration).
    cool thanks and yes I remembered the constant of integration
 
 
 
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