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Math Help - Integral again

  1. #1
    Member javax's Avatar
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    Integral again

    Need help with this integral!
    \int{\frac{dx}{[(x-1)^2+3]\sqrt{3-(x-1)^2}}}

    I start by substituting x-1=t but don't know how to go further...
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  2. #2
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    try u = 3- t
    I like dealing with easier roots
    Might help here too
    Last edited by jbpellerin; September 14th 2008 at 06:18 PM. Reason: woops
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  3. #3
    Member javax's Avatar
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    Quote Originally Posted by jbpellerin View Post
    try u = 3- t
    I like dealing with easier roots
    Might help here too
    But here we have 3-t^2...I don't see it working...
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  4. #4
    Super Member 11rdc11's Avatar
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    \int{\frac{dt}{[t^2+3]\sqrt{3-t^2}}}<br />

    Divide by radical 3

    \frac{1}{\sqrt{3}}\int{\frac{dt}{[t^2+3]\sqrt{1-(\frac{t}{\sqrt{3}})^2}}}<br />

    Now sub

    \frac{t}{\sqrt3} =sin\theta

    and use the identity

    1- sin^2\theta = cos^2\theta


    \frac{1}{\sqrt{3}}\int \frac{\sqrt{3}cos\theta(d\theta)}{[3sin^2\theta +3]cos\theta} = \int \frac{cos\theta(d\theta)}{[3sin^2\theta +3]cos\theta} = \int\frac{cos\theta(d\theta)}{(3-3cos^2\theta+3)cos\theta}

    = \int\frac{cos\theta(d\theta)}{(6 -3cos^2\theta)cos\theta}= \int\frac{d\theta}{6-3cos^2\theta} = \int \frac{d\theta}{3(sin^2\theta + 1)}
    Last edited by 11rdc11; September 14th 2008 at 09:15 PM. Reason: correction thanks javax
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  5. #5
    Member javax's Avatar
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    but by dividing with \sqrt{3} wouldn't it be
     <br />
\frac{1}{\sqrt{3}} \int{\frac{dt}{[t^2+3]\sqrt{1-(\frac{t}{\color{red}{\sqrt{3}}}})^2}}<br />
    or I messed it?
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  6. #6
    Super Member 11rdc11's Avatar
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    \int\frac{\frac{dt}{\sqrt{3}}}{\frac{[t^2+3]\sqrt{3-t^2}}{\sqrt{3}}}

    Now remember

    \frac{\sqrt{3-t^2}}{\sqrt{3}} = \sqrt{\frac{3-t^2}{3}} = \sqrt{1-\bigg(\frac{t}{\sqrt{3}}\bigg)^2}
    Last edited by 11rdc11; September 14th 2008 at 07:57 PM. Reason: Correction thanks javax
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  7. #7
    Member javax's Avatar
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    yes man
    \sqrt{\frac{3-t^2}{3}} = \sqrt{1-\frac{t^2}{3}} and not \sqrt{1-\left(\frac{t}{3}\right)^2}
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  8. #8
    Super Member 11rdc11's Avatar
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    Thanks for pointing that out. I can't believe I did that, I'm ashamed lol. Ok let me go fix my mistake.
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  9. #9
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    Here's my approach:

    \int{\frac{dx}{[(x-1)^2+3]\sqrt{3-(x-1)^2}}}

    First substitution to make things look neat:
    u = x-1

    du = dx

    \int{\frac{du}{[u^2+3]\sqrt{3-u^2}}}

    Second substitution:
    t = \frac{1}{u^2+3}

    dt = -\frac{2u}{(u^2+3)^2}

    3-u^2 = \frac{6t-1}{t}

    u = \sqrt{\frac{1-3t}{t}}

    Replacing, we get:

    -\frac{1}{2} \int \frac{1}{t \sqrt{\frac{1-3t}{t}} \sqrt{\frac{6t-1}{t}}}

    = -\frac{1}{2} \int \frac{1}{(\sqrt{1-3t})(\sqrt{6t-1})}

    = -\frac{1}{2} \int \frac{1}{\sqrt{-18t^2+9t-1}}

    By completing the square, we get:
    = -\frac{\sqrt{8}}{2} \int \frac{1}{\sqrt{1-(12t-3)^2}}

    Done!

    After some backsubbing and simplification:

    = -\frac{\sqrt{2}}{12} \arcsin{\left(\frac{-3x^2+6x}{(x-1)^2+3}\right)} + C
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