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Math Help - integral

  1. #1
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    integral

    Hey,

    Unfortunately I am not able to solve the following integral:
    ∫1/(SQRT(4-64(x)^2))dx

    Can somebody help me?
    Thanks
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  2. #2
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    Use \int \frac{1}{\sqrt{1-u^2}}~du = \sin^{-1}u+C
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  3. #3
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    Thanks for your reply. I had that formula in mind as well, but don't really have an idea how to apply it to this exercise.
    Last edited by naomi; January 23rd 2010 at 01:30 PM.
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  4. #4
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    Quote Originally Posted by naomi View Post
    Thanks for your reply. I had that formula in mind as well, but don't not really have an idea how to apply it to this exercise.
    4-64x^2 = 4(1-16x^2)

    Therefore use the sub u=4x
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  5. #5
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    Use

    4-64x^2=4(1-16x^2)

    \sqrt{4-64x^2}=\sqrt{4}\sqrt{1-16x^2}=\sqrt{4}\sqrt{1-(4x)^2}
    Last edited by Archie Meade; January 23rd 2010 at 01:23 PM.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    Use

    4-64x^2=4(1-16x^2)

    \sqrt{4-64x^2}= \sqrt{2}\sqrt{1-16x^2}=\sqrt{2}\sqrt{1-(4x)^2}

    \sqrt{4-64x^2}= \sqrt{4}\sqrt{1-16x^2}=2\sqrt{1-(4x)^2} \neq\sqrt{2}\sqrt{1-16x^2}
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  7. #7
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    Quote Originally Posted by naomi View Post
    Hey,

    Unfortunately I am not able to solve the following integral:
    ∫1/(SQRT(4-64(x)^2))dx

    Can somebody help me?
    Thanks
    EDIT: I missread the post

    What we have is of the form \int\frac{1}{\sqrt{a^2-u^2}}du=\arcsin\frac{u}{a}+C .

    In this case, it is easy to see that a=2 and u=8x.
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  8. #8
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    Quote Originally Posted by VonNemo19 View Post
    \int\sqrt{4-64x^2}dx
    I read the original post to be

    \int\frac{1}{\sqrt{4-64x^2}}dx
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  9. #9
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    u=4x, du=4dx
    ∫1/(2sqrt(1-(u)^2))dx
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  10. #10
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    so when a=2 and u = 8x then,
    ∫1/(sqrt(2^2-(8x)^2))dx
    How can I put this into the following format: arcsin u/a +c?
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  11. #11
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    Hi naomi,

    you can also use trigonometric substitutions.

    One way is to draw a right-angled triangle with
    perpendicular sides

    \sqrt{4-64x^2}\ and 8x.

    The hypotenuse of this triangle is 2, since 2^2-(8x)^2=4-64x^2

    Then, with the acute angle opposite the side of length 8x, we get

    Sin\theta=\frac{8x}{2}=4x

    \frac{d}{dx}Sin\theta=4

    dx=\frac{1}{4}dSin\theta

    hence

    \int{\frac{1}{\sqrt{4-64x^2}}}dx=\frac{1}{8}\int{\frac{1}{Cos\theta}}dSi  n\theta
    Last edited by Archie Meade; January 23rd 2010 at 02:56 PM.
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  12. #12
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    Quote Originally Posted by naomi View Post
    so when a=2 and u = 8x then,
    ∫1/(sqrt(2^2-(8x)^2))dx
    How can I put this into the following format: arcsin u/a +c?
    Well, a=2 and u=8x so \int\frac{1}{\sqrt{(2)^2-(8x)^2}}dx=\arcsin\frac{8x}{2}+C=\arcsin(4x)+C
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  13. #13
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    @ VonNemo19:
    Shouldn't the answer be: 1/8 arcsin(4x)+c instead of arcsin(4x)+c?

    @ archie meade

    Thanks for your explanation about the application of trigonometric substitutions to this exercise. I didn't know I could use this for this exercise. Trigonometric substitutions are quite new to me.
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  14. #14
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    Hi naomi,

    yes, the trig substitutions can be tried later in your development.

    It's best for you to master the inverse sine solution
    recommended by the other contributors earlier.

    \int{\frac{1}{\sqrt{a^2-y^2}}}dx

    dx needs to be changed to dy

    a^2=4,\ a=2

    y^2=64x^2=(8x)^2,\ y=8x
    Last edited by Archie Meade; January 24th 2010 at 03:06 AM. Reason: typo
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  15. #15
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    Quote Originally Posted by naomi View Post
    Hey,

    Unfortunately I am not able to solve the following integral:
    ∫1/(SQRT(4-64(x)^2))dx

    Can somebody help me?
    Thanks
    I would use trigonometric substitution in this case.

    You have \int{\frac{1}{\sqrt{4 - 64x^2}}\,dx} = \int{\frac{1}{\sqrt{64\left(\frac{1}{16} - x^2\right)}}\,dx}

     = \int{\frac{1}{8\sqrt{\frac{1}{16} - x^2}}\,dx}

     = \frac{1}{8}\int{\frac{1}{\sqrt{\left(\frac{1}{4}\r  ight)^2 - x^2}}\,dx}.


    Now make the substitution x = \frac{1}{4}\sin{\theta} so that dx = \frac{1}{4}\cos{\theta}\,d\theta.


    So the integral becomes

    \frac{1}{8}\int{\frac{1}{\sqrt{\left(\frac{1}{4}\r  ight)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2}}\,\frac{1}  {4}\cos{\theta}\,d\theta}

    = \frac{1}{32}\int{\frac{1}{\frac{1}{4}\sqrt{1 - \sin^2{\theta}}}\,\cos{\theta}\,d\theta}

     = \frac{1}{8}\int{\frac{\cos{\theta}}{\sqrt{1 - \sin^2{\theta}}}\,d\theta}

     = \frac{1}{8}\int{\frac{\cos{\theta}}{\sqrt{\cos^2{\  theta}}}\,d\theta}

     = \frac{1}{8}\int{\frac{\cos{\theta}}{\cos{\theta}}\  ,d\theta}

     = \frac{1}{8}\int{1\,d\theta}

     = \frac{1}{8}\theta + C.


    Now, remembering that you made the original substitution x = \frac{1}{4}\sin{\theta}

    4x = \sin{\theta}

    \theta = \arcsin{4x}.


    So your final answer is \frac{1}{8}\arcsin{4x} + C.
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