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    questions 3 and 4 - any ideas?

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    What have you tried?
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    (Original post by quirksy)
    questions 3 and 4 - any ideas?

    different techniques required for most of them ...
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    (Original post by zetamcfc)
    What have you tried?
    (Original post by TeeEm)
    different techniques required for most of them ...
    well i'm currently attempting 3a, and trying integration by substitution...

    substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.
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    (Original post by quirksy)
    well i'm currently attempting 3a, and trying integration by substitution...

    substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.
    you mean1/(9-9sin2(u)) 3 cos(u) du

    and also a square root

    factorize the 9 under the square root ...
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    (Original post by quirksy)
    well i'm currently attempting 3a, and trying integration by substitution...

    substituted in x=3sin(u) to get integral of 1/(3-3sin(u)) 3 cos(u) du.
    If you substituted x=3sin(u) then x^2 wouldn't be 3sin(u). You should have integral 1/sqrt[(9-9sin^2(u)] . 3cos(u) du

    You have said that sqrt(9-9x^2)=3-3x which isn't true.
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    too many of us here ...
    I am the weakest link...
    goodbye
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    (Original post by TeeEm)
    you mean1/(9-9sin2(u)) 3 cos(u) du

    and also a square root

    factorize the 9 under the square root ...
    But I square rooted the 9 - 9sin2(u) to get 3 - 3sin(u)?
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    (Original post by quirksy)
    But I square rooted the 9 - 9sin2(u) to get 3 - 3sin(u)?
    please read post 6 and post 7
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    (Original post by quirksy)
    But I square rooted the 9 - 9sin2(u) to get 3 - 3sin(u)?
    That's not how square roots work!

    Are you saying that \sqrt{a^2 - b^2} = a - b, because that isn't true at all!!
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    (Original post by Protoxylic)
    If you substituted x=3sin(u) then x^2 wouldn't be 3sin(u). You should have integral 1/sqrt[(9-9sin^2(u)] . 3cos(u) du

    You have said that sqrt(9-9x^2)=3-3x which isn't true.
    oh right, thank you! is this correct?

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    (Original post by quirksy)
    oh right, thank you! is this correct?

    Correct
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    i find these ones interesting, so i`ll indicate them:

    4).

    b) use the substitution: u=x^{2}+2x

    d) use the relation: \displaystyle \sinA \cosB = \frac{1}{2}  ( \sin(A+B) + \sin(A-B)) to reduce to a more integrable form.

    f) divide out into partial fraction then use logs to integrate.

    I) use integration by parts (or tabular integration - it`s quicker)

    j) expand out, and reduce by using the identity: - \cot^{2}(x)=1- \csc^{2}(x)

    k) use partial fractions.
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    It really is quite simple, have you tried using examsolutions.net the guy doing the videos is extremely helpful.
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    (Original post by quirksy)
    oh right, thank you! is this correct?

    No

    You need to change your limits
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    (Original post by buxtonarmy)
    How'd you solve 4c?
    You turn it into A + B/(2x+1)
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    (Original post by buxtonarmy)
    What does that mean?
    It means what it says!

    You can rewrite the integrand as the sum of a constant and a fraction which has a number on top and a linear factor on the bottom.

    You know how to integrate a constant and you should know how to integrate something of the form A/(ax + b) where A, a and b are constants
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    (Original post by TenOfThem)
    No

    You need to change your limits
    to what?

    edit: don't worry I've got it now, thanks!
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    (Original post by TeeEm)
    too many of us here ...
    I am the weakest link...
    goodbye
    PRSOM
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    (Original post by buxtonarmy)
    So is fractional division required?
    You can do polynomial division if you wish

    A number of methods would work
 
 
 
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