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Math Help - Integration by substitution help

  1. #1
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    Integration by substitution help

    Use the substitution x = sin\theta to find the exact value of;

     \int^{\frac{1}{2}}_0 \frac{1}{( {1-x^{2}})^\frac{3}{2}}

     \int \frac{1}{(1-sin^{2}\theta)^{\frac{3}{2}}}

     \frac{dx}{d\theta} = cos\theta

    so when I put this all together I get ;

     \int \frac{1}{(cos^{2}\theta)}^{\frac{3}{2}} cos\theta d\theta

    and there does not see to be much I can do from here, cant seem to cancel the cosines,




    and the 3/2 power is suppose to be at the bottom, of the fraction. Cant seem to do it on latex
    Last edited by Tweety; June 17th 2010 at 12:55 PM.
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  2. #2
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    Try re-posting this one for now without LaTeX.
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  3. #3
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    Or with LaTeX, now. You have to manually put in the
    Code:
    [tex]
    tags around the LaTeX source code, I think.
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  4. #4
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    Here... unfortunately, though, you may have to wait for the latex to work.

    Did you forget that (x^m)^(n/p) = x^{(mn)/p}, i.e. {(cosx)^2}^(3/2) = (cosx)^{(3*2)/2} = cos^3{x}? Full solution here, but you may have to wait for the latex to work.

    We have [LaTeX ERROR: Convert failed] If we let (as suggested) [LaTeX ERROR: Convert failed] , then [LaTeX ERROR: Convert failed] When [LaTeX ERROR: Convert failed] , [LaTeX ERROR: Convert failed] ; similarly, when [LaTeX ERROR: Convert failed] , [LaTeX ERROR: Convert failed] . Then we have [LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed] [LaTeX ERROR: Convert failed]
    Last edited by TheCoffeeMachine; June 17th 2010 at 07:15 PM.
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  5. #5
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    Quote Originally Posted by TheCoffeeMachine View Post
    Full solution here, but you may have to wait for the latex to work.
    Used this page to make an image, useful for now, click on image for full size:

    Integration by substitution help-int_substitution.png

    For fast results, used search-and-replace

    [tex] -> }
    [/tex] -> \text{

    and then check the beginning and end to make sure no missing or extra \text{ or }
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  6. #6
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    Here is a trick , substitute x = 1/t we have dx = -dt/t^2 and it becomes


    int_{oo}^{2} (-dt/t^2)/( 1 - 1/t^2 )^(3/2)

    = int_{2}^{oo} [dt/t^2(t^3)] /( t^2 -1 )^(3/2)

    = int_{2}^{oo} (t dt)(t^2 - 1)^(-3/2)

    = [ (t^2-1)^(-1/2) (-2) / 2 ]_{2}^{oo}

    = - [ 1/sqrt(t^2-1) ]_{2}^{oo}

    = - 0 + 1/sqrt(2^2-1) = 1/sqrt(3)
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  7. #7
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    Integration by substitution help-lll.gif
    Quote Originally Posted by undefined View Post
    Used this page to make an image, useful for now, click on image for full size:

    Click image for larger version. 

Name:	int_substitution.png 
Views:	17 
Size:	8.5 KB 
ID:	17894

    For fast results, used search-and-replace

    [tex] -> }
    [/tex] -> \text{

    and then check the beginning and end to make sure no missing or extra \text{ or }
    Cool, thanks! Recreated the picture, with corrections of typos that I've (no wonder) made earlier:

    Attached Thumbnails Attached Thumbnails Integration by substitution help-99.gif  
    Last edited by TheCoffeeMachine; June 17th 2010 at 07:50 PM.
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Cool, thanks! Recreated the picture, with corrections of typos that I've (no wonder) made earlier:
    You're welcome!

    How did you get the inline image to appear full size? I see that the code is

    [IMG]http://www.mathhelpforum.com/math-help/attachment.php?attachmentid=17895&d=1276832168[/IMG]

    but what steps did you use to produce it?
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  9. #9
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    Quote Originally Posted by undefined View Post
    You're welcome!

    How did you get the inline image to appear full size? I see that the code is

    [IMG]http://www.mathhelpforum.com/math-help/attachment.php?attachmentid=17895&d=1276832168[/IMG]

    but what steps did you use to produce it?
    After uploading, click on the attachment (where it says Attached Thumbnails), and then it will take you to the image url. (When you click to view the image you posted earlier in full size, clicking again will give you the URL).

    EDIT: Just realised that might not have been clear.

    After uploading, I clicked 'preview post' and clicked on the attachment, then it took me to here. I copied http://www.mathhelpforum.com/math-he...5&d=1276832168 from the address bar and went back to replying, clicked on the image icon in the tool bar, and pasted it on the little box that appears, and the picture appeared full.
    Last edited by TheCoffeeMachine; June 18th 2010 at 07:48 AM.
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