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    (Original post by Zaphod77)
    I'll be honest, I can't think of a way... I'd just stick to sec^2(x), a lot easier! I was trying to manipulate it, but I'm not sure you can... Sorry!
    Do you think you would be asked to integrate 4(3x+2)^-5... or is that not in our spec?

    Also when you have an improper fraction(top heavy) is there a easier way to workout what A and B will go over as numerators to the denominators?
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    (Original post by nm786)
    expand the brackets and you get at-at^2 - so yes use product rule
    what would be the u and v?
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    (Original post by Chris-69)
    And for that type you would be A + B/(whatever) + C/(whatever) ?

    uhm im not sure what you mean but once i know it is a top heavy fraction I would just work out the A and b/c normally
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    (Original post by otrivine)
    hi

    integrating sin3xcosx dx
    Use factor formulae from C3...

    (Original post by otrivine)
    differenitate


    at(1-t) ?

    use product rule?
    You didn't say with respect to what... but I assume with respect to "t"... & "a" is a constant

    I would personally open out the brackets....

    at - at^2

    dy/dt = a - 2at
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    (Original post by Story)
    Do you think you would be asked to integrate 4(3x+2)^-5... or is that not in our spec?

    Also when you have an improper fraction(top heavy) is there a easier way to workout what A and B will go over as numerators to the denominators?
    There's no reason why they can't ask us that.
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    (Original post by otrivine)
    what would be the u and v?
    You are thinking to hard now. That's C1 differentiation, except 'a' is 'the number'.
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    (Original post by Fortitude)
    Ok first look at page 93, Example 6c.

    1. You can see that sin3xcosx is in the addition formula sin (A+-B) = sinAcosB +- cosAsinB. A is obvs 3x & B is x.
    2. So you can have sin (3x + x) = sin3xcosx + cos3xsinx
    3. But you need to get sin3xcosx
    4. You can also have sin (3x - x) = sin3xcosx - cos3xsinx
    5. We need to get just sin3xcosx on it's own so in order to do that, you can tell that we need to add the equations: (sin3xcosx + cos3xsinx) + (sin3xcosx - cos3xsinx) so we end up with 2sin3xcosx.
    6. You need to also therefore add the sin (3x+x) + sin(3x-x) so you get sin 4x + sin2x = 2sin3xcosx
    7. Hence now just divide by 2 so you get 1/2(sin4x + sin2x) = sin3xcosx
    8. Just integrate (sin4x + sin2x) & then x1/2 & subs limits in!!!

    actually instead of doing the books way,

    can I not use sub method to assume u=sin3x? and do it from there or will it over complicate things
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    (Original post by MasterYi)
    You are thinking to hard now. That's C1 differentiation, except 'a' is 'the number'.
    But the guy said use prodct?
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    not looking forward to it at all.
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    (Original post by usycool1)
    Because Edexcel told us we need to keep the restrictions and not all international countries do a different paper. :nah:
    I see! Are you doing the paper/
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    (Original post by hawraaj313)
    not looking forward to it at all.

    WAR CRY!
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    (Original post by otrivine)
    But the guy said use prodct?
    If you used product rule for the unexpanded version you would end up with the same answer.

    Guy meant product for the unexpanded version. I think.
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    (Original post by kronca)
    uhm im not sure what you mean but once i know it is a top heavy fraction I would just work out the A and b/c normally
    I mean if it's top heavy, would you do for example:

    A + B/(x+1) + C(x+2)
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    is there any specific soloman press paper worth doing that may help in exam or shell I just do all of them?
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    (Original post by otrivine)
    actually instead of doing the books way,

    can I not use sub method to assume u=sin3x? and do it from there or will it over complicate things
    That would still be complicated as you'd have to cosx in terms of u??:confused:
    Practise some questions in the book to get the hang of it.
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    (Original post by Frankster)
    There's no reason why they can't ask us that.
    Would it be -1/3(3x+1)^-4 ?
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    (Original post by Story)
    Do you think you would be asked to integrate 4(3x+2)^-5... or is that not in our spec?
    they can definitely ask you to integrate that btw
    you can usually do ones like that pretty easy by inspection :P
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    (Original post by Story)
    Do you think you would be asked to integrate 4(3x+2)^-5... or is that not in our spec?

    Also when you have an improper fraction(top heavy) is there a easier way to workout what A and B will go over as numerators to the denominators?
    You could be, you can use a substitution to do that or do it by inspection. Your second question, if you have an improper fraction I tend to divide through (long division) so then you get a number and a remainder, which will be over the original denominator. The remainder can then be split into partial fractions, which you can do either through comparing coefficients when you've set up the A and B fraction or through eliminating them (e.g. if the fraction included A(x+2) you can let x=-2 to eliminate A and just leave B). Does that answer your question at all?
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    (Original post by MasterYi)
    If you used product rule for the unexpanded version you would end up with the same answer.

    Guy meant product for the unexpanded version. I think.
    I see,

    question 6) ?

    http://www.school-portal.co.uk/Group...urceId=3992067
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    (Original post by otrivine)
    I see! Are you doing the paper/
    Yup, I am :yep:
 
 
 
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