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    Help needed with integration! please give me the steps on integrating this. tan(2x - π/6) with lower limit 0 and upper limit π/12. how can i get the answer 1/4 ln3.
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    When differentiating y = 2x x 4x, I get 22x(3ln2), yet the book says the answer is 23x(3ln2).

    Can anyone explain how to get the right answer?
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    can they ever ask us to rotate something more/less than 2pi radians? any examples of questions or explanations on what they would ask and how to go about it?

    (Original post by TeddyBasherz)
    Does anyone have a link to the jan 2013 c4 paper? The booklet version not the version on examsolutions? Thanks
    http://eiewebvip.edexcel.org.uk/Repo...s_20130307.pdf
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    hi how do I integrate Name:  ImageUploadedByStudent Room1371405121.475193.jpg
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Size:  138.0 KB

    thanks

    ps that is 2^2-2x -1

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    (Original post by task)
    hi how do I integrate Name:  ImageUploadedByStudent Room1371405121.475193.jpg
Views: 123
Size:  138.0 KB

    thanks

    ps that is 2^2-2x -1

    Posted from TSR Mobile

    have you tried using sub method? so assume u=2-2x ?
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    (Original post by task)
    hi how do I integrate Name:  ImageUploadedByStudent Room1371405121.475193.jpg
Views: 123
Size:  138.0 KB

    thanks

    ps that is 2^2-2x -1

    Posted from TSR Mobile
    rewrite as  4(2^{-2x}) and should be simple from there
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    (Original post by otrivine)
    have you tried using sub method? so assume u=2-2x ?
    no I haven't ill give that a go now, thanks
    didn't even think of that oops


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    (Original post by TDL)
    rewrite as  4(2^{-2x}) and should be simple from there
    do you think if you use a sub method? is okay
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    (Original post by TDL)
    rewrite as  4(2^{-2x}) and should be simple from there
    sorry could you explain how you got that?


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    Does anyone know how to integrate 2^x???? Is it always 2^x/ln2? (say: if y=4.5^t does the integral = 4.5^t/ln4.5)?
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    http://www.examsolutions.net/a-level...1&solution=4.4


    why and where did he get the 2 from?
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    where can i get past replacement papers?
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    (Original post by task)
    sorry could you explain how you got that?


    Posted from TSR Mobile
    sure

     2^{2-2x} = 2^2 \times 2^{-2x} = 4 \times 2^{-2x}

    full solution

    Spoiler:
    Show

     4(2^{-2x}) = 4e^{ln(2^{-2x})}

    take the derivative of that and you'll get  -8(2^{-2x})ln(2) notice that this can be written as -2(4(2^{-2x}))ln(2) which then tells you that  \displaystyle \int (4(2^{-2x})) \ dx  = - \dfrac{1}{2ln(2)} (4(2^{-2x}))


    hope this helps
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    (Original post by ITomI)
    u = cosx
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    (Original post by task)
    no I haven't ill give that a go now, thanks
    didn't even think of that oops


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    did it work with sub method??
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    (Original post by silentriver)
    Help needed with integration! please give me the steps on integrating this. tan(2x - π/6) with lower limit 0 and upper limit π/12. how can i get the answer 1/4 ln3.
    I think your answer is wrong btw. I checked on my calculator and got 1/2 ln(root3/2). Anyway, I wrote it as the integral of sin(2x-pi/6)/cos(2x-pi/6), and then manipulated it to get it in the form: integral of f'(x)/f(x). You'll get an ln function for your result, and then sub your limits.
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    (Original post by physicso)
    Does anyone know how to integrate 2^x???? Is it always 2^x/ln2? (say: if y=4.5^t does the integral = 4.5^t/ln4.5)?
    This should help (it's similar)

    (Original post by TDL)
    sure

     2^{2-2x} = 2^2 \times 2^{-2x} = 4 \times 2^{-2x}

    full solution

    Spoiler:
    Show

     4(2^{-2x}) = 4e^{ln(2^{-2x})}

    take the derivative of that and you'll get  -8(2^{-2x})ln(2) notice that this can be written as -2(4(2^{-2x}))ln(2) which then tells you that  \displaystyle \int (4(2^{-2x})) \ dx  = - \dfrac{1}{2ln(2)} (4(2^{-2x}))


    hope this helps
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    (Original post by otrivine)
    did it work with sub method??
    I couldn't do it but probably because I suck at C4 haha


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    (Original post by TDL)
    sure

     2^{2-2x} = 2^2 \times 2^{-2x} = 4 \times 2^{-2x}

    full solution

    Spoiler:
    Show

     4(2^{-2x}) = 4e^{ln(2^{-2x})}

    take the derivative of that and you'll get  -8(2^{-2x})ln(2) notice that this can be written as -2(4(2^{-2x}))ln(2) which then tells you that  \displaystyle \int (4(2^{-2x})) \ dx  = - \dfrac{1}{2ln(2)} (4(2^{-2x}))


    hope this helps
    thank you so much, hope nothing like this comes up!


    Posted from TSR Mobile
 
 
 
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