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    (Original post by Rayquaza)
    Yup.
    did you use substitution method for the 1/(1-y^2) ?
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    (Original post by otrivine)
    did you use substitution method for the 1/(1-y^2) ?
    No...
    Think that's where I went wrong. I'll try it now.
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    (Original post by usycool1)
    The one about volumes?

    You just need to solve the differential equation.
    Hi

    how would you integrate

    question 6)a)

    http://www.school-portal.co.uk/Group...urceId=3992058

    I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
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    (Original post by Rayquaza)
    No...
    Think that's where I went wrong. I'll try it now.
    Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?
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    (Original post by otrivine)
    Hi

    how would you integrate

    question 6)a)

    http://www.school-portal.co.uk/Group...urceId=3992058

    I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
    Check your formula booklet and you should be able to see something that looks similar in it. If not just ask and I'll help you out
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    (Original post by Rayquaza)
    Attachment 227150
    This is my working. Try to make sense of it, its a bit all over the place.


    Posted from TSR Mobile
    wouldn't you do partial fractions rather than reverse chain rule?
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    (Original post by usycool1)
    The one about volumes?

    You just need to solve the differential equation.
    Yeah, that one. I kind of forgot all about differential equations! Thanks!
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    (Original post by otrivine)
    Hi

    how would you integrate

    question 6)a)

    http://www.school-portal.co.uk/Group...urceId=3992058

    I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
    Factor formulae?
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    (Original post by MathsNerd1)
    Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?
    Hi

    so do you mean
    use


    oh yes, true,
    I can use partial fractions and integrate separately, cause I was trying the sub method but looks confusing to integrate
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    (Original post by otrivine)
    Hi

    how would you integrate

    question 6)a)

    http://www.school-portal.co.uk/Group...urceId=3992058

    I tried to use the Sin(A+B) formula but it looks really confusing and horrible to work wit
    Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.
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    (Original post by tiny hobbit)
    Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.
    Hi

    So we will never be asked this question?
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    (Original post by masryboy94)
    wouldn't you do partial fractions rather than reverse chain rule?

    (Original post by MathsNerd1)
    Surely you don't need a substitution just expand the denominator as its the difference of two squares, so can't you separate by partial fractions and then integrate each term separately?
    So as my partial fractions I got 1/(1-y) and 1/(y+1)
    Correct?
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    (Original post by usycool1)
    Factor formulae?
    can you show me please how to start of with
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    (Original post by Rayquaza)
    So as my partial fractions I got 1/(1-y) and 1/(y+1)
    Correct?
    yepp and now integrate
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    (Original post by Rayquaza)
    So as my partial fractions I got 1/(1-y) and 1/(y+1)
    Correct?
    That's almost correct but I got a 1/2(1+y) + 1/2(1-y) can you see where you went wrong? Ignore this as this would only be for if you're separating 1/(1-y^2) and not the 2/(1-y^2) that you have done
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    (Original post by Rayquaza)
    Attachment 227145
    Question 4e)


    Answer:
    4e)

    Attachment 227146

    Getting a weird decimal for '+c'
    Help?



    Posted from TSR Mobile
    you method just before integration is correct: split the fraction of 2/(1-y^2) up into partial fractions. you should get 1/(1-y) and 1/(1+y). Integrate (1/(1-y) + 1/(1+y)) dy = integration 1/(1+x) dx After integration you will be left with ln(1+y) - ln(1-y) = ln(1+x) + c sub into the above x=5 and y=0.5 rearrange to get c=ln(0.5) rewrite the whole equation as; ln(1+y) - ln(1-y) = ln(1+x) + ln(0.5) now put it all together; ln((1+y)/(1-y)) = ln ((1+x)/2) sorry it s bit messy!!!! where ever I wrote 0.5 think of it as 1/2
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    how would i integrate this  \int sec^3 3x
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    (Original post by masryboy94)
    yepp and now integrate
    Are you sure about that? Try expanding your partial fractions and you'll notice that it won't equal 1/(1-Y^2) unless I'm just being stupid Ignore me I've seen what you've done already
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    (Original post by tiny hobbit)
    Although it's on the Solomon papers, it's not on the Edexcel spec, even though the factor formulae are given in the formula book.
    It isn't on the spec? then why are there questions on it in the edexcel text book?
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    (Original post by MathsNerd1)
    Are you sure about that? Try expanding your partial fractions and you'll notice that it won't equal 1/(1-Y^2) unless I'm just being stupid
    Your partial fractions are correct
 
 
 
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