Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

STEP 1994 Question

Announcements Posted on
Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
  1. Offline

    ReputationRep:
    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:	75 
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. Offline

    ReputationRep:
    (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:	75 
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.
  3. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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:	75 
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. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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?
  7. Offline

    ReputationRep:
    (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?
  8. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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.
  10. Offline

    ReputationRep:
    (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. Offline

    ReputationRep:
    (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

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 3, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Would you consider Clearing if you missed your offer?
Results and Clearing

Results are coming...

No sweat. Here's all you need to make sure you're ready

new on tsr

What's your life ambition?

Graduating, travelling, owning a business?

Study resources
x

Think you'll be in clearing or adjustment?

Hear direct from unis that want to talk to you

Get email alerts for university course places that match your subjects and grades. Just let us know what you're studying.

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