Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    In the Solomon paper L for C4 (OCR), theres the following question:

    Evaluate
    (with limits b = pi/3, and a = 0)

    ∫ Sin2x.Cosx dx

    In the markscheme they've solved it as:

    ∫2Sinx.Cos²x dx

    Which i assume they got from expanding Sin2x = 2SinxCosx

    And they then jump to:

    [(-2/3).Cos³x] with the limits from above.

    I assume what they've done here is use a substitution of u=Cosx, so du/dx = -Sinx, so the integral becomes:

    ∫(2Sinx.u²) / -sinx = ∫ -2u² du = [ (-2/3)u³ ] = [ (-2/3)Cos³x ]

    However i confused myself here thinking it wasnt correct to integrate Cos²x like this, as i thought that when integrating cos²x and sin²x, i had to use the double angle identities.

    So is what i've done here the correct way to solve it, even though i didnt think it was correct maths to integrate cos²x this way? And whats the best way for me to identify which integration method to use on this sort of question? (I thought i had to integrate this by parts when i first saw it).
    Offline

    10
    ReputationRep:
    You weren't integrating cos²x here -- it was cos²x.sinx, which is very different.

    The markscheme used the "function and its derivative" rule, which is essentially, as you spotted, using the substitution u=cosx with du/dx=-sinx.

    If you did have cos²x by itself, then you would've needed to use the appropriate double-angle identity for cos2x.
    Offline

    2
    ReputationRep:
    Indeed, spotting a function and derivative is handy for integrating. And a substitution in this case is fine. All you've got to practise is being able to spot such cases!
    • Thread Starter
    Offline

    0
    ReputationRep:
    Ok, thanks for the help.
    Offline

    13
    ReputationRep:
    Alternatively,
    Using sin(A)cos(B) = 0.5[ sin(A+B) + sin(A-B) ]
    Sin(2x)Cos(x) = 0.5[ sin(3x) + sin(x) ]
    Which integrates to
    0.5[ -(1/3)cos(3x) - cos(x) ] + c
    Limits:
    0.5[ (1/3) - (1/2) + (1/3) + 1 ]
    = 7/12
    Offline

    0
    ReputationRep:
    (Original post by Migraine)
    ∫2Sinx.Cos²x dx
    Which i assume they got from expanding Sin2x = 2SinxCosx
    And they then jump to:
    [(-2/3).Cos³x] with the limits from above.
    its done my a method called "recognition" - you merely look at the integral and work backwards saying "what function must i differentiate wto get the integral"

    wol ah,

    Maths made easy

    regarding method - what tehy tell you to use - if its an easy one, they'll explect you dto do it..ones like that they may well give you a substitution because the majority of people cant do recognition

    pk
    Offline

    0
    ReputationRep:
    Is that the same as "standard patterns"?
    Offline

    13
    ReputationRep:
    (Original post by seankhn)
    Is that the same as "standard patterns"?
    You what :confused:
    Offline

    15
    ReputationRep:
    I would say it's done by knowing the chain rule that g(f(x)) differentiates to

    g'(f(x)) f'(x)
 
 
 
Turn on thread page Beta
Updated: June 17, 2005
The home of Results and Clearing

1,068

people online now

1,567,000

students helped last year

University open days

  1. SAE Institute
    Animation, Audio, Film, Games, Music, Business, Web Further education
    Thu, 16 Aug '18
  2. Bournemouth University
    Clearing Open Day Undergraduate
    Fri, 17 Aug '18
  3. University of Bolton
    Undergraduate Open Day Undergraduate
    Fri, 17 Aug '18
Poll
Do you want your parents to be with you when you collect your A-level results?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

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