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Math Help - Final integrals

  1. #1
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    Final integrals

    I want to compare my answers to your's?
    I will be glad if someone will solve them?
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  2. #2
    TD!
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    If you only want to compare solutions, I'm giving my answers for now:

    4) 1/6

    5) \frac{2\cdot5^{{\sqrt{1 + 3\,x}}}}{3\cdot{\log (5)}^2}\left({\sqrt{1 + 3\,x}}\,\log (5)-1 \right)

    10) doesn't converge

    11) \frac{\pi}{8 \cdot \sqrt{5}}

    If you have something else, I can post more details.
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  3. #3
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    Hello, sbsite!

    This is what I got for #4 . . .


    4)\;\;\int^2_1\frac{dx}{(x^2 - 2x + 4)^{\frac{3}{2}}}

    The denominator has: . x^2 - 2x + 1 + 3\:=\:(x-1)^2 + 3

    Let x - 1 \:=\:\sqrt{3}\tan\theta\quad\Rightarrow\quad dx \:=\:\sqrt{3}\sec^2\theta\,d\theta

    When x = 1\!:\;\;\sqrt{3}\tan\theta = 0\quad\Rightarrow\quad\theta = 0
    When x = 2\!:\;\;\sqrt{3}\tan\theta = 1\quad\Rightarrow\quad\tan\theta = \frac{1}{\sqrt{3}}\quad\Rightarrow\quad\theta = \frac{\pi}{6}


    Substitute: . \int^{\frac{\pi}{6}}_0 \frac{\sqrt{3}\sec^2\theta\,d\theta}{(3\sec^2\thet  a)^{\frac{3}{2}}} \;=\;\int^{\frac{\pi}{6}}_0\frac{\sqrt{3}\sec^2\th  eta\,d\theta}{3\sqrt{3}\sec^3\theta} \;=\;\frac{1}{3}\int^{\frac{\pi}{6}}_0\frac{d\thet  a}{\sec\theta} = \;\frac{1}{3}\int^{\frac{\pi}{6}}_0\cos\theta\,d\t  heta

    And we have: . \frac{1}{3}\sin\theta\,\bigg|^{\frac{\pi}{6}}_0 \;=\;\frac{1}{3}\sin\frac{\pi}{6} - \frac{1}{3}\sin0 \;=\;<br />
\frac{1}{3}\left(\frac{1}{2}\right) - \frac{1}{3}(0) \;=\;\boxed{\frac{1}{6}}

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