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Math Help - Integration ...... again

  1. #1
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    Integration ...... again

    Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this ---> .

    1. \int\frac{4x}{x^2 +3} where the substitution is u = x^2 + 3

    and

    2. \int\cos^3(x) where the substitution is s = \sin(x)

    Thanks for any help
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  2. #2
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    Quote Originally Posted by Beard View Post
    Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this ---> .

    1. \int\frac{4x}{x^2 +3} where the substitution is u = x^2 + 3

    and

    2. \int\cos^3(x) where the substitution is s = \sin(x)

    Thanks for any help
    First things first, remember to put in your "dx", when writing the integral. That thing is crucial to subsitution.

    1. \int\frac{4x}{x^2 +3} dx where the substitution is u = x^2 + 3

    and

    2. \int\cos^3(x)dx where the substitution is s = \sin(x)

    1.  \frac{du}{dx} = 2x

    Hence  xdx = \frac{1}{2}du . Just replace xdx with that, and replace x^2+3 with u.

    2.  \frac{ds}{dx} = cos(x)dx

     ds = cos(x)dx

     I = \int\cos^3(x)dx = \int cos^2(x).cos(x)dx

    You need to write  cos^2(x) in terms of sin(x) so you can use your substitution, s.
    Last edited by Mush; December 29th 2008 at 08:11 AM.
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  3. #3
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    [quote=Beard;241407]Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this ---> .

    1. \int\frac{4x}{x^2 +3} where the substitution is u = x^2 + 3
    If we make the sub u=x^{2}+3, \;\ du=2xdx, \;\ \frac{du}{2}=xdx. We make the subs and get:

    2\int\frac{1}{u}du

    Now, go to town.
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  4. #4
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    Quote Originally Posted by Beard View Post
    Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this ---> .

    1. \int\frac{4x}{x^2 +3} where the substitution is u = x^2 + 3

    and

    2. \int\cos^3(x) where the substitution is s = \sin(x)

    Thanks for any help
    Rewrite \int\frac{4x}{x^2 +3}~dx as 2\int\frac{2x}{x^2 +3}~dx

    Sub u = x^2 + 3 \implies du = 2x~dx. This switches the integral to:

    2\int\frac{1}{u}~du

    Which should be a standard form you should pick up right away.

    2. Subbing s = \sin{x} \implies ds = \cos{x}~dx switches the integral to:

    \int \cos^2{x}~ds

    Notice how we managed to cancel one factor of x. Using the identity \cos^2{x} = 1-\sin^2{x}, the integral becomes:

    \int (1-s^2)~ds

    \ldots
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  5. #5
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    Mush gave you some very good hints


    The solution for your first problem is:

      \int \frac{4x}{x^2+3} dx

    u:=x^2 + 3
    => u' = 2x

     u' = \frac{du}{dx} => dx = \frac{du}{u'}


     \int \frac{4x}{x^2+3} dx  = \int \frac{4x}{u}*\frac{du}{u'}

      = \int \frac{4x}{u}*\frac{du}{2x} =  = \int \frac{2}{u}*du

    I hope you know the answer now!

    Quote Originally Posted by Mush View Post
    First things first, remember to put in your "dx", when writing the integral. That thing is crucial to subsitution.

    1. \int\frac{4x}{x^2 +3} dx where the substitution is u = x^2 + 3

    and

    2. \int\cos^3(x)dx where the substitution is s = \sin(x)

    1.  \frac{du}{dx} = 2x

    Hence  xdx = \frac{1}{2}du . Just replace xdx with that, and replace x^2+3 with u.

    2.  \frac{ds}{dx} = cos(x)dx
     ds = cos(x)dx

     I = \int\cos^3(x) = \int cos^2(x).cos(x)dx

    You can write cos^2(x) in terms of sin(x).
    you should use:
    sin^2+cos^2 = 1
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  6. #6
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    Thanks but I had already done those processes with both of them, its just getting past that point that I find difficult.

    i.e. the first one

    \frac{4x}{x^2 +3} and u = x^2 +3

    then,as you were saying,

    \frac{du}{2} = x.dx which would then mean

    \frac{2}{u}. Carrying on, if I was then to integrate this I would get

    2\ln(x^2 +3) +c.
    Is this correct?

    then the second one I had done what you said before but I just get sort of tangled up.

    \cos^3(x).dx

    where s = sin(x) due to the trigonometric identity

    \cos^2(x) = 1 -\sin^2(x) becomes

    (1-s^2)\cos(x) or \cos - s^2\cos(x)

    also \frac{ds}{dx} = cos(x).
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  7. #7
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    Hey mate

    Quote Originally Posted by Beard View Post
    Thanks but I had already done those processes with both of them, its just getting past that point that I find difficult.

    i.e. the first one

    \frac{4x}{x^2 +3} and u = x^2 +3

    then,as you were saying,

    \frac{du}{2} = x.dx which would then mean

    \frac{2}{u}. Carrying on, if I was then to integrate this I would get

    2\ln(x^2 +3) +c.
    Is this correct?
    Yup, the first one is correct: 2 ln(x^2 +3) +c

    Well done, Beard!

    Quote Originally Posted by Beard View Post
    then the second one I had done what you said before but I just get sort of tangled up.

    \cos^3(x).dx

    where s = sin(x) due to the trigonometric identity

    \cos^2(x) = 1 -\sin^2(x) becomes

    (1-s^2)\cos(x) or \cos - s^2\cos(x)

    also \frac{ds}{dx} = cos(x).
    Well, it is

     \int cos^3  dx = \int cos^2(x) cos(x) dx

    s:= sin(x)
    => s' = cos(x)

      s' = \frac{ds}{dx} => dx = ds/s' = ds / cos(x)

      \int cos^2(x) cos(x) dx = \int cos^2(x) *cos(x) \frac{ds}{cosx}

      = \int cos^2(x) du = \int (1-sin^2(x) ) ds = \int (1-s^2) ds


    because s = sin(x)
    Think about it!
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  8. #8
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    And *ding* the lightbulb goes on, for the second one anyway. As it finishes:

    \int(1 - s^2).ds becoming s-\frac{s^3}{3} +c

    Although you said I was right for the first question, my books disagrees, and although I would love to believe the book is wrong it generally isn't and it gives the answer of:

    -\frac{1}{(x^2 +3)^2} +c
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  9. #9
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    Quote Originally Posted by Beard View Post
    And *ding* the lightbulb goes on, for the second one anyway. As it finishes:

    \int(1 - s^2).ds becoming s-\frac{s^3}{3} +c

    Although you said I was right for the first question, my books disagrees, and although I would love to believe the book is wrong it generally isn't and it gives the answer of:

    -\frac{1}{(x^2 +3)^2} +c
    It is
    \int\frac{4x}{x^2 +3} \not= -\frac{1}{(x^2 +3)^2} +c

    But why is that?

    Because

     [-\frac{1}{(x^2 +3)^2} +c]' = \frac{4x}{(x^2 + 3)^3}


    Edit: \int(1 - s^2).ds becoming s-\frac{s^3}{3} +c

    Thats the right answer
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  10. #10
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    Quote Originally Posted by Rapha View Post
    It is
    \int\frac{4x}{x^2 +3} \not= -\frac{1}{(x^2 +3)^2} +c

    But why is that?

    Because

     [-\frac{1}{(x^2 +3)^2} +c]' = \frac{4x}{(x^2 + 3)^3}
    I concur. Either the book is wrong, or else Mr. Beard has misread it.
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  11. #11
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    All settled, here, I know, but here's a couple of pics...








    As usual, straight continuous lines diff/anti-diff with respect to x, straight dashed lines with respect to the dashed balloon expression.

    Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers
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