Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

Step ii 2007 q6

Announcements Posted on
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    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---
    • 41 followers
    Offline

    ReputationRep:
    (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.
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    Great! Thanks ---
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    (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 ---
    • 41 followers
    Offline

    ReputationRep:
    (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.
    • 14 followers
    Offline

    ReputationRep:
    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)
    • Thread Starter
    • 0 followers
    Offline

    ReputationRep:
    Thank you very much ---

    I've also fixed my typo.

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: June 4, 2012
New on TSR

Student in a Million awards

All the results from our prize-giving night

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.