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Thread: Integral Questions

  1. #1
    Senior Member polymerase's Avatar
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    Integral Questions

    How do u integrate:

    $\displaystyle
    \int {x sec^{ 2 } x \,dx}
    $

    $\displaystyle
    \int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
    $

    $\displaystyle
    \int_{0}^{1} {x 3^{ x } \,dx}
    $

    $\displaystyle
    \int \dfrac{ln \frac{1}{x}}{x^{3}}
    $

    Help on any would be greatly appreciated .
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by polymerase View Post
    How do u integrate:

    $\displaystyle
    \int {x sec^{ 2 } x \,dx}
    $

    $\displaystyle
    \int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
    $

    $\displaystyle
    \int_{0}^{1} {x 3^{ x } \,dx}
    $

    $\displaystyle
    \int \dfrac{ln \frac{1}{x}}{x^{3}}
    $

    Help on any would be greatly appreciated .
    You need to use integration by parts. General rule is to put u equal to the bit
    of the integrand that you can differentiate most easily (of the part that you can't integrate).

    So in the first you can differentiate $\displaystyle x$ easily and integrate $\displaystyle \sec^2(x)$ easily, so you would use $\displaystyle u=x, dv=\sec^2(x)$

    RonL
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You need to use integration by parts. General rule is to put u equal to the bit
    of the integrand that you can differentiate most easily (of the part that you can't integrate).

    So in the first you can differentiate $\displaystyle x$ easily and integrate $\displaystyle \sec^2(x)$ easily, so you would use $\displaystyle u=x, dv=\sec^2(x)$

    RonL
    For the 2nd question i yet u = x and dv = 1/x and when i integrate i end up with:

    = xlnx - \int ln x dx
    = x lnx - (xlnx -x)
    = x
    but thats not the right answer. any recommendation?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    $\displaystyle
    \int_{1}^{e} \dfrac{ln \frac{1}{x}}{x}
    $
    Remember, your $\displaystyle u dv$ must equal the integrand!

    Let
    $\displaystyle u = ln \left ( \frac{1}{x} \right ) \implies du = -\frac{dx}{x}$
    $\displaystyle dv = \frac{dx}{x} \implies v = ln(x)$

    So
    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

    Do integration by parts on the second integral:
    $\displaystyle \int \frac{ln(x)}{x} dx$
    Let
    $\displaystyle u = ln(x) \implies du = \frac{dx}{x}$
    $\displaystyle dv = \frac{dx}{x} \implies v = ln(x)$

    So
    $\displaystyle \int \frac{ln(x)}{x} dx = ln^2(x) - \int \frac{ln(x)}{x}dx$

    Thus
    $\displaystyle 2 \int \frac{ln(x)}{x} dx = ln^2(x)$

    $\displaystyle \int \frac{ln(x)}{x} dx = \frac{ln^2(x)}{2}$

    So back to the original integral:
    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \frac{ln^2(x)}{2} + C$

    I'll let you put the limits in.

    -Dan
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  5. #5
    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    Remember, your $\displaystyle u dv$ must equal the integrand!

    Let
    $\displaystyle u = ln \left ( \frac{1}{x} \right ) \implies du = -\frac{dx}{x}$
    $\displaystyle dv = \frac{dx}{x} \implies v = ln(x)$

    So
    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

    Do integration by parts on the second integral:
    $\displaystyle \int \frac{ln(x)}{x} dx$
    Let
    $\displaystyle u = ln(x) \implies du = \frac{dx}{x}$
    $\displaystyle dv = \frac{dx}{x} \implies v = ln(x)$

    So
    $\displaystyle \int \frac{ln(x)}{x} dx = ln^2(x) - \int \frac{ln(x)}{x}dx$

    Thus
    $\displaystyle 2 \int \frac{ln(x)}{x} dx = ln^2(x)$

    $\displaystyle \int \frac{ln(x)}{x} dx = \frac{ln^2(x)}{2}$

    So back to the original integral:
    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \int \frac{ln(x)}{x} dx$

    $\displaystyle \int \frac{ln \frac{1}{x}}{x}dx = ln(x) \cdot ln \left ( \frac{1}{x} \right ) + \frac{ln^2(x)}{2} + C$

    I'll let you put the limits in.

    -Dan
    Thank You
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  6. #6
    Eater of Worlds
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    Parts:

    $\displaystyle \int{x\cdot{3^{x}}}dx$

    Let $\displaystyle u=x, \;\ dv=3^{x}, \;\ du=dx, \;\ v=\frac{3^{x}}{ln(3)}$

    $\displaystyle uv-\int{vdu}$

    $\displaystyle \frac{x\cdot{3^{x}}}{ln(3)}-\frac{1}{ln(3)}\int{3^{x}}dx$

    $\displaystyle \frac{x\cdot{3^{x}}}{ln(3)}-\frac{3^{x}}{(ln(3))^{2}}=\boxed{\frac{3^{x}}{ln(3 )}\left[x-\frac{1}{ln(3)}\right]}$
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  7. #7
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    Hello, polymerase!

    Here's the second one . . .


    $\displaystyle \int_{1}^{e} \frac{\ln(\frac{1}{x})}{x}\,dx$
    The numerator is: .$\displaystyle \ln\left(\frac{1}{x}\right) \:=\:\ln\left(x^{-1}\right) \:=\:-\ln(x)$

    The integral becomes: .$\displaystyle -\int\ln(x)\cdot\frac{dx}{x}$

    . . Let: $\displaystyle u \,=\,\ln x\quad\Rightarrow\quad du \,=\,\frac{dx}{x}$

    Substitute: .$\displaystyle -\int u\,du \:=\:-\frac{1}{2}u^2 + C$

    Back-substitute: .$\displaystyle -\frac{1}{2}\left[\ln(x)\right]^2 + C$

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