STEP 1994 Question

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  1. shanban's Avatar
    1 03-06-2012  13:51
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    STEP 1994 Question
    So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

    It's q 4ii) on 1994 STEP I (attached below)

    Click image for larger version.  Name: stepq.jpg  Views: 66  Size: 46.1 KB  ID: 153549

    I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

    Am I just not using the method correctly?

  2. Blutooth's Avatar
    2 03-06-2012  13:54
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    Re: STEP 1994 Question
    (Original post by shanban)
    So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

    It's q 4ii) on 1994 STEP I (attached below)

    Click image for larger version.  Name: stepq.jpg  Views: 66  Size: 46.1 KB  ID: 153549

    I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

    Am I just not using the method correctly?

    You need to use the sub cosalpha =k to get rid of the sinalpha at the top.


    Eidt: forget what I said was talking nonsense.
    Last edited by Blutooth; 03-06-2012 at 13:56.
  3. shanban's Avatar
    3 03-06-2012  14:11
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    Re: STEP 1994 Question
    (Original post by Blutooth)
    You need to use the sub cosalpha =k to get rid of the sinalpha at the top.


    Eidt: forget what I said was talking nonsense.
    I was just wondering why it wouldn't work without a trig sub...but thanks anyway!
  4. M1K3's Avatar
    4 03-06-2012  14:29
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    Re: STEP 1994 Question
    (Original post by shanban)
    So I've been struggling on this q for a while...not because I don't know how to do it, but because I can't seem to work out what I'm doing wrong!

    It's q 4ii) on 1994 STEP I (attached below)

    Click image for larger version.  Name: stepq.jpg  Views: 66  Size: 46.1 KB  ID: 153549

    I'm using the complete-the-square method but this way I don't have a square root above the denominator of 1/(1-k^2)

    Am I just not using the method correctly?

    Are we on about the first part of ii)?

    We complete the square to get \displaystyle\int \dfrac{1}{(x-k)^2 + (1-k^2)}\ dx are u alrite with this or is it something else?
  5. shanban's Avatar
    5 03-06-2012  16:01
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    Re: STEP 1994 Question
    (Original post by M1K3)
    Are we on about the first part of ii)?

    We complete the square to get \displaystyle\int \dfrac{1}{(x-k)^2 + (1-k^2)}\ dx are u alrite with this or is it something else?
    Yeah I got that, and then I tried to make it into an integral which would then give something along the lines of arctan...but the constant I took out of the integral didn't have a square root where it should
  6. TheJ0ker's Avatar
    6 03-06-2012  16:07
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    Re: STEP 1994 Question
    (Original post by shanban)
    Yeah I got that, and then I tried to make it into an integral which would then give something along the lines of arctan...but the constant I took out of the integral didn't have a square root where it should
    Have you tried a half tan substitution? I think that is what the question is looking for.

    EDIT: in fact if you have \displaystyle\int \frac{1}{(x-k)^2 + (\sqrt{1-k^2})^2} dx then can you see what to do?
    Last edited by TheJ0ker; 03-06-2012 at 16:16.
  7. shanban's Avatar
    7 03-06-2012  16:19
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    Re: STEP 1994 Question
    (Original post by TheJ0ker)
    Have you tried a half tan substitution? I think that is what the question is looking for.

    EDIT: in fact if you have \displaystyle\int \frac{1}{(x-k)^2 + (\sqrt{1-k^2})^2} dx then can you see what to do?
    This is what I've done:

    \dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx = \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C

    I don't understand why you'd put the sqrt in?
    Last edited by shanban; 03-06-2012 at 16:21.
  8. TheJ0ker's Avatar
    8 03-06-2012  16:23
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    Re: STEP 1994 Question
    (Original post by shanban)
    This is what I've done:

    \dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx = \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C

    I don't understand why you'd put the sqrt in?
    Woah why are you taking stuff out of the integral? Do you know what the standard result is? It's on the formula book FP3 part.
  9. M1K3's Avatar
    9 03-06-2012  21:28
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    Re: STEP 1994 Question
    (Original post by shanban)
    This is what I've done:

    \dfrac{1}{1-k^2} \int \frac{1}{(\frac{x-k}{\sqrt {1-k^2}})^2 + 1} dx = \dfrac{1}{1-k^2} arctan \dfrac{x-k}{\sqrt{1-k^2}} + C

    I don't understand why you'd put the sqrt in?
    As theJoker said, this is a standard integral.

    But your way is fine, the reason why your method has not worked is because i think you mite have done the integration slightly wrong, well, just differentiate your answer and i think you'll understand.

    Remember, you need to divide by the differential of \dfrac{x-k}{\sqrt{1-k^2}} , yeah, chain rule backwards.
    Last edited by M1K3; 03-06-2012 at 21:30.
  10. shanban's Avatar
    10 03-06-2012  22:26
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    Re: STEP 1994 Question
    (Original post by M1K3)
    As theJoker said, this is a standard integral.

    But your way is fine, the reason why your method has not worked is because i think you mite have done the integration slightly wrong, well, just differentiate your answer and i think you'll understand.

    Remember, you need to divide by the differential of \dfrac{x-k}{\sqrt{1-k^2}} , yeah, chain rule backwards.
    oh! I just realised I took the differential of \dfrac{x-k}{\sqrt{1-k^2}} to be 1 when it obviously isn't! I'll try again and see if I can get it now...

    thanks for your help!
  11. M1K3's Avatar
    11 03-06-2012  23:38
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    Re: STEP 1994 Question
    (Original post by shanban)
    oh! I just realised I took the differential of \dfrac{x-k}{\sqrt{1-k^2}} to be 1 when it obviously isn't! I'll try again and see if I can get it now...

    thanks for your help!
    np
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