Step ii 2007 q6

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  1. adrienne_om's Avatar
    1 01-06-2012  18:29
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    Step ii 2007 q6
    I'm not sure if math questions for STEP are meant to be posted in separate posts. So I apologise if I haven't gotten it right. Feel free to merge this post.

    Actual question:


    \text{Differentiate} \ln(x + \sqrt{3 + x^2}) \text{and} x\sqrtsqrt{3 + x^2} .
    \text{Hence find} \int \sqrt{3 + x^2} dx .

    I'm asking about chewwy's solution to Q6 in STEP II 2007:

    http://www.thestudentroom.co.uk/atta...4&d=1183161671

    \int \sqrt{3 + x^2} dx

    = \displaystyle \int \sqrt{3 + x^2} + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} dx

    First Question --- How would you see this trick --- to add \displaystyle \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} to the desired integral?

    I know the question asks us to differentiate the functions but I still don't see this trick.

    Then use \frac{d}{dx}\sqrt{3 + x^2} = ... to get

    \int \sqrt{3 + x^2} dx + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}}

    \displaystyle = x\sqrt{3 + x^2}- \left \int \frac{x^2}{\sqrt{3 + x^2}} \right

    Second Question --- I know how to integrate -\int \frac{x^2}{\sqrt{3 + x^2}} by using the substitution x = atan(u) . But the STEP solution says that no calculus work is needed for this part --- the integration of \int \sqrt{3 + x^2} dx . I don't see the derivatives in the first part help me to get \displaystyle -\int \frac{x^2}{\sqrt{3 + x^2}}?

    Thanks a lot---
    Last edited by adrienne_om; 04-06-2012 at 19:12.
  2. Farhan.Hanif93's Avatar
    2 01-06-2012  20:50
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    Re: Step ii 2007 q6
    (Original post by adrienne_om)
    I'm not sure if math questions for STEP are meant to be posted in separate posts. So I apologise if I haven't gotten it right. Feel free to merge this post.

    Actual question:


    \text{Differentiate} \ln(x + \sqrt{3 + x^2}) \text{and} x\sqrtsqrt{3 + x^2} .
    \text{Hence find} \int \sqrt{3 + x^2} dx .

    I'm asking about chewwy's solution to Q6 in STEP II 2007:

    http://www.thestudentroom.co.uk/atta...4&d=1183161671

    \int \sqrt{3 + x^2} dx

    = \displaystyle \int \sqrt{3 + x^2} + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} dx

    First Question --- How would you see this trick --- to add \displaystyle \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}} to the desired integral?

    I know the question asks us to differentiate the functions but I still don't see this trick.

    Then use \frac{d}{dx}\sqrt{3 + x^2} = ... to get

    \int \sqrt{3 + x^2} dx + \frac{x^2}{\sqrt{3 + x^2}} - \frac{x^2}{\sqrt{3 + x^2}}

    \displaystyle = x\sqrt{3 + x^2}- \left \int \frac{x^2}{\sqrt{3 + x^2}} \right
    He seems to have used a less simplified form of the derivative of x\sqrt{3+x^2}. Namely, by the product rule, \dfrac{d}{dx}[x\sqrt{3+x^2}] = \sqrt{3+x^2} + \dfrac{x^2}{\sqrt{3+x^2}}. As soon as he saw that, it's obvious that he'll be able to evaluate an integral similar to that so he went for the addition-subtraction trick.

    Second Question --- I know how to integrate -\int \frac{x^2}{\sqrt{3 + x^2}} by using the substitution x = atan(u) . But the STEP solution says that no calculus work is needed for this part --- the integration of \int \sqrt{3 + x^2} dx . I don't see the derivatives in the first part help me to get \displaystyle -\int \frac{x^2}{\sqrt{3 + x^2}}?

    Thanks a lot---
    He then noted a different form of the above derivative: \dfrac{d}{dx}[x\sqrt{3+x^2}] = \dfrac{3+2x^2}{\sqrt{3+x^2}} (1)

    and also:
    \dfrac{d}{dx}[\ln(x+\sqrt{3+x^2})] = \dfrac{1}{\sqrt{3+x^2}} (2)

    Furthermore:
    \dfrac{x^2}{\sqrt{3+x^2}} = \dfrac{1}{2}\left(\dfrac{3+2x^2} {\sqrt{3+x^2}} - \dfrac{3}{\sqrt{3+x^2}}\right)
    = \dfrac{1}{2}\left(\dfrac{d}{dx}[x\sqrt{3+x^2}] -3\dfrac{d}{dx}[\ln(x+\sqrt{3+x^2})]\right)

    by (1) and (2), which is easy to integrate. He made a mistake in the last line of working - There's a small sign error.

    I have to say, I think this is the long way of doing it. When I did the question, I noted that:

    \dfrac{d}{dx}(x\sqrt{3+x^2}) = 2\sqrt{3+x^2} - \dfrac{3}{\sqrt{3+x^2}}
    \iff 2\sqrt{3+x^2} = \dfrac{d}{dx}(x\sqrt{3+x^2}) + \dfrac{3}{\sqrt{3+x^2}}

    Using (2), the RHS is easy to integrate and you're done.
    Last edited by Farhan.Hanif93; 01-06-2012 at 21:13. Reason: LaTeX being buggy.
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  3. adrienne_om's Avatar
    3 02-06-2012  21:12
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    Re: Step ii 2007 q6
    Great! Thanks ---
  4. adrienne_om's Avatar
    4 03-06-2012  04:21
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    Re: Step ii 2007 q6
    (Original post by Farhan.Hanif93)
    I have to say, I think this is the long way of doing it. When I did the question, I noted that:

    \dfrac{d}{dx}(x\sqrt{3+x^2}) = 2\sqrt{3+x^2} - \dfrac{3}{\sqrt{3+x^2}}
    \iff 2\sqrt{3+x^2} = \dfrac{d}{dx}(x\sqrt{3+x^2}) + \dfrac{3}{\sqrt{3+x^2}}

    Using (2), the RHS is easy to integrate and you're done.
    actually, one last question ---

    How did you see the above? It wasn't and is still not obvious to me. The only way I see it is ---

    \dfrac{3 + 2x^2}{\sqrt{3 + x^2}} = \dfrac{2(x^2 + 3) - 3}{\sqrt{x^2 + 3}} = \dfrac{2(x^2 + 3)}{\sqrt{x^2 + 3}} - \dfrac{3}{\sqrt{x^2 + 3}}

    and it looks like only to be a very clever algebra trick ---

    Thanks again ---
    Last edited by adrienne_om; 04-06-2012 at 19:11.
  5. Farhan.Hanif93's Avatar
    5 03-06-2012  04:44
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    Re: Step ii 2007 q6
    (Original post by adrienne_om)
    actually, one last question ---

    How did you see the above? It wasn't and is still not obvious to me. The only way I see it is ---

    \dfrac{2 + 3x^2}{\sqrt{3 + x^2}} = \dfrac{3(x^2 + 2) - 3}{\sqrt{x^2 + 3}} = \dfrac{3(x^2 + 2)}{\sqrt{x^2 + 3}} - \dfrac{3}{\sqrt{x^2 + 3}}

    and it looks like only to be a very clever algebra trick ---

    Thanks again ---
    I think you've made a small typo at the start; it should be \dfrac{3+2x^2}{\sqrt{x^2+3}}. My reasoning behind it was that I wanted to rewrite this derivative in terms of \sqrt{3+x^2} AND a constant multiple of the same or other derivative. It was clear to me that \dfrac{2 + 3x^2}{\sqrt{3 + x^2}} can be rewritten in that form so I did the working and it just fell out.
    Last edited by Farhan.Hanif93; 03-06-2012 at 04:45.
  6. Intriguing Alias's Avatar
    6 03-06-2012  08:13
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    Re: Step ii 2007 q6
    How I did it was as follows:

    \dfrac{d}{dx} \left( \ln(x+\sqrt{x^2+3}) \right) = \frac{1}{3} \left( \sqrt{x^2+3} - \dfrac{x^2}{\sqrt{x^2+3}} \right)

    and \dfrac{d}{dx} \left( x \sqrt{3 + x^2} \right) = \sqrt{x^2 + 3} + \dfrac{x^2}{\sqrt{x^2+3}}


    Now we can easily see that \displaystyle \int \sqrt{x^2 + 3} = \dfrac{1}{2} \left( x \sqrt{3 + x^2} + 3 \ln(x+\sqrt{x^2+3}) \right)
  7. adrienne_om's Avatar
    7 04-06-2012  19:12
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    Re: Step ii 2007 q6
    Thank you very much ---

    I've also fixed my typo.
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