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Math Help - integral

  1. #16
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by naomi View Post
    @ VonNemo19:
    Shouldn't the answer be: 1/8 arcsin(4x)+c instead of arcsin(4x)+c?
    yes it should. Check out the hundred other posts here, I apologize.
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  2. #17
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    Quote Originally Posted by VonNemo19 View Post
    No, but I would like to hear what makes you ask that question.
    Yes it would. You have forgotten to take into account that \frac{du}{dx} = 8.

    So you will need to turn the integral into

    \frac{1}{8}\int{8f\left(u(x)\right)\,dx}.
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  3. #18
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    hi naomi,

    following on from Prove It's good post,

    in terms of the substitution you are working with,
    we have

    \int{\frac{1}{\sqrt{4-64x^2}}}dx

    If we write this as

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

    then

    y^2=64x^2=(8x)^2

    y=8x

    \frac{dy}{dx}=8

    dy=8dx

    dx=\frac{dy}{8}

    hence

    \int{\frac{1}{\sqrt{4-64x^2}}}dx=\int{\frac{1}{\sqrt{2^2-(8x)^2}}}dx=\int{\frac{1}{\sqrt{2^2-y^2}}}\frac{dy}{8}

    =\frac{1}{8}\int{\frac{1}{\sqrt{2^2-y^2}}}dy=\frac{1}{8}arcsin(4x)+C
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  4. #19
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    thank you all for your help.
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  5. #20
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    Hi Prove it,

    your trigonometric approach is almost clear to me. There is one thing which I don't completely get.
    How do you get from this step sqrt((1/4)^2-(1/4sin (theta))^2) to 1/4 sqrt(1-sin^2(theta))?
    Last edited by naomi; January 26th 2010 at 01:01 PM.
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  6. #21
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    Quote Originally Posted by naomi View Post
    Hi Prove it,

    your trigonometric approach is almost clear to me. There is one thing which I don't completely get.
    How do you get from this step sqrt((1/4)^2-(1/4sin (theta))^2) to 1/4 sqrt(1-sin^2(theta))?
    \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\sin{\theta}\right)^2} = \sqrt{\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^2\sin^2{\theta}}

     = \sqrt{\left(\frac{1}{4}\right)^2(1 - \sin^2{\theta})}

     = \sqrt{\left(\frac{1}{4}\right)^2}\sqrt{1 - \sin^2{\theta}}

     = \frac{1}{4}\sqrt{1 - \sin^2{\theta}}

     = \frac{1}{4}\sqrt{\cos^2{\theta}}

     = \frac{1}{4}\cos{\theta}.


    Step 1: Take the square through all the factors.

    Step 2: Take out a common factor of \left(\frac{1}{4}\right)^2.
    Last edited by Prove It; January 27th 2010 at 08:34 PM.
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  7. #22
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    Hi Prove it

    Could you explain how you get from

    sqrt(-(1/4)^2((sin^2)(theta)))
    to
    sqrt((1-sin^2(theta)))
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  8. #23
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    Check my edit.
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  9. #24
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    Thank you.
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