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Math Help - Integral and series help.

  1. #1
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    Integral and series help.

    1. Integral of (x(ln(x))/(square root of (x^2)-1) dx

    2. Integral of cos^2(1+ln(x))/x dx

    3. Evaluate the following sum: (SERIES QUESTION)

    n=1 to infinity (3^(n+2))/(2^(2n-1))
    Last edited by mr fantastic; December 15th 2010 at 08:46 PM. Reason: Deleted begging etc. from title.
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  2. #2
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    Quote Originally Posted by devil10078 View Post
    1. Integral of (x(ln(x))/(square root of (x^2)-1) dx

    2. Integral of cos^2(1+ln(x))/x dx

    3. Evaluate the following sum: (SERIES QUESTION)

    n=1 to infinity (3^(n+2))/(2^(2n-1))
    #3.

    \displaystyle 2^{(2n-1)}={{2^{2n}}\over{2}}={{4^{n}}\over{2}}.

    \displaystyle 3^{(n+2)}=9\left(3^n\right).

    So  \displaystyle {{3^{(n+2)}}\over{2^{(2n-1)}}}= {{9\left(3^n\right)}\over{{{4^{n}}\over{2}}}}= {{18\left(3^n\right)}\over{4^{n}}}.

    Therefore,  \displaystyle \sum_{n=1}^\infty}{{3^{(n+2)}}\over{2^{(2n-1)}}}= \sum_{n=1}^\infty}18{{3^n}\over{4^{n}}}

    Can you take it from there?
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by devil10078 View Post
    ]
    2. Integral of cos^2(1+ln(x))/x dx
    Let  u = 1+\ln x . What does the integral become?
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  4. #4
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    \displaystyle \int \frac{\cos^2(\ln{x}+1)}{x} \, dx<br />

    u = \ln{x} + 1

    \displaystyle du = \frac{1}{x} \, dx

    substitute ...

    \displaystyle \int \cos^2{u} \, du = \frac{1}{2} \int 1 + \cos(2u) \, du<br />

    integrate then back sub ...
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  5. #5
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    Hello, devil10078!

    I pray that these are not on your exam . . .


    \displaystyle (1)\;\;I \;=\;\int\frac{x \ln x }{\sqrt{x^2-1}}\,dx

    Integrate by parts:

    . . \begin{array}{ccccccccc}<br />
u &=& \ln x && dv &=& x(x^2-1)^{\frac{1}{2}}\,dx \\<br />
du &=& \dfrac{dx}{x} && v &=& \frac{1}{3}(x^2-1)^{\frac{3}{2}} \end{array}

    \displaystyle I \;=\;\tfrac{1}{3}(x^2-1)^{\frac{3}{2}}\ln x - \tfrac{1}{3}\int \frac{(x^2-1)^{\frac{3}{2}}}{x}\,dx


    The new integral reguires Trig Substitution:

    . . x \,=\,\sec\theta \quad\Rightarrow\quad dx \,=\,\sec\theta\tan\theta\,d\theta

    Good Luck!

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