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    OK - for some reason I can't get these ones - i know I should, but my mind just ain't focusing.

    1) Integral of x/(x-1)^1/2 dx

    2) Integral of cos 2x sin x dx

    For number 2, I've been using the substitution u = sin x, so that dx will then equal du/cos x. But which identity do I use for cos 2x?
    I'd appreciate some help, so I can get back on track. Cheers.
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    (Original post by foowise)
    OK - for some reason I can't get these ones - i know I should, but my mind just ain't focusing.

    1) Integral of x/(x-1)^1/2 dx

    2) Integral of cos 2x sin x dx

    For number 2, I've been using the substitution u = sin x, so that dx will then equal du/cos x. But which identity do I use for cos 2x?
    I'd appreciate some help, so I can get back on track. Cheers.
    1) You'll have to use integration by parts. differentiate 'x' so it dissapears and integrate '(x-1)^-1/2'

    2) Don't use a substitution, just turn cos2X into (2Cos^2x - 1), sub it in and you're there.
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    OK - these might seem dumb - but could you please post your methods?
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    For question number 2, i would use:

    cos2x = 2cos^2x - 1

    so you would get INT (2cos^2x - 1)sinx
    Expanding brackets would give you:

    INT 2cox^2xsinx - sinx

    for integral of 2cox^2xsinx, use substitution method where u=cos x

    for integral of sinx, use normal trig integration.

    Hope this helps
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    Here's my method on an attatchment.
    Attached Files
  1. File Type: doc Maths solutions.doc (23.5 KB, 78 views)
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    (Original post by foowise)
    OK - for some reason I can't get these ones - i know I should, but my mind just ain't focusing.

    1) Integral of x/(x-1)^1/2 dx

    2) Integral of cos 2x sin x dx

    For number 2, I've been using the substitution u = sin x, so that dx will then equal du/cos x. But which identity do I use for cos 2x?
    I'd appreciate some help, so I can get back on track. Cheers.
    If you want to take a slightly different approach, take the trigonometric identity of sinA-sinB=2cos[(A+B)/2]sin[(A-B)/2] (it's in the formula book)
    Select two appropriate values for A and B (let A=3x and B=x) and we have:
    sin3x-sinx=2cos2xsinx
    And so cos2xsinx=o.5sin3x-0.5sinx which is easy to integrate.

    1000th post
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    OK - thanx guys.

    For number 1 - i used the fact that d/dx(cos^3 x) is -3 cos^2 x sin x to integrate the 2 sin x cos^2 x bit. I'm assuming that is ok (?), since the answer came out right.

    But Mysticmin - apparently the answer for number 2 is 2/3(x+2)(x-1)^(1/2)? It's number 26 in Ex 4E in the P3 textbook. I can't see where they get the (x+2) bracket from.
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    (Original post by foowise)
    OK - thanx guys.

    For number 1 - i used the fact that d/dx(cos^3 x) is -3 cos^2 x sin x to integrate the 2 sin x cos^2 x bit. I'm assuming that is ok (?), since the answer came out right.

    But Mysticmin - apparently the answer for number 2 is 2/3(x+2)(x-1)^(1/2)? It's number 26 in Ex 4E in the P3 textbook.
    Yeah, the answer you quoted was the same as the one I got, you take a
    (x-1)^1/2 outside the bracket as it's a common factor of both terms and you get that answer.
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    OK thanx! God, I am a right royal retard - and to think I'm taking the exam tomorrow. But first, I need a break.

    Cheers again.
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    (Original post by foowise)
    OK - for some reason I can't get these ones - i know I should, but my mind just ain't focusing.

    1) Integral of x/(x-1)^1/2 dx

    2) Integral of cos 2x sin x dx

    For number 2, I've been using the substitution u = sin x, so that dx will then equal du/cos x. But which identity do I use for cos 2x?
    I'd appreciate some help, so I can get back on track. Cheers.
    {x/x-1^1/2 . dx
    let U = x - 1
    du/dx = 1 , dx = 1
    x = u + 1

    { u + 1/ u^1/2 .dx = { u/u^1/2 + 1/U^1/2 .dx

    { u + u^-1/2 .dx = u^2/2 + u^1/2/2 + c

    (x - 1)^2/2 + (x - 1)^1/2/2 + c
    Im not sure though.
 
 
 
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