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Thread: integration technique

  1. #1
    Junior Member erich22's Avatar
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    integration technique

    I encounter 2 problems when I explain integration technique to my student.
    Could someone help me?
    Thx alot.

    \int{x^5*[(4-3x^4)]^1/3}=
    \int{\frac{e^x}{2+e^(2x)}}=
    all with respect to x.


    GBU
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  2. #2
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    $\displaystyle \int{\frac{e^x}{2 + e^{2x}}\,dx} = \int{\frac{e^x\,dx}{2 + (e^x)^2}}$.

    Let $\displaystyle u = e^x$ so that $\displaystyle du= e^x\,dx$ and the integral becomes

    $\displaystyle \int{\frac{du}{2 + u^2}}$.


    Now you have to use a trigonometric substitution. Let $\displaystyle u = \sqrt{2}\tan{\theta}$ so that $\displaystyle \theta = \arctan{\left(\frac{\sqrt{2}u}{2}\right)}$ and $\displaystyle du = \sqrt{2}\sec^2{\theta}\,d\theta$. The integral becomes

    $\displaystyle \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2 + (\sqrt{2}\tan{\theta})^2}}$

    $\displaystyle = \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2 + 2\tan^2{\theta}}}$

    $\displaystyle = \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2(1 + \tan^2{\theta})}}$

    $\displaystyle = \int{\frac{\sqrt{2}\sec^2{\theta}\,d\theta}{2\sec^ 2{\theta}}}$

    $\displaystyle = \frac{\sqrt{2}}{2}\int{1\,d\theta}$

    $\displaystyle = \frac{\sqrt{2}}{2}\theta + C$

    $\displaystyle = \frac{\sqrt{2}}{2}\arctan{\left(\frac{\sqrt{2}u}{2 }\right)} + C$

    $\displaystyle = \frac{\sqrt{2}}{2}\arctan{\left(\frac{\sqrt{2}e^x} {2}\right)} + C$.
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  3. #3
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    I would use integration by parts on the first integral...

    $\displaystyle \int{x^5(4 - 3x^4)^{\frac{1}{3}}\,dx} = -\frac{1}{12}\int{x^2\left(-12x^3\right)(4 - 3x^4)^{\frac{1}{3}}\,dx}$.


    Now let $\displaystyle u = x^2$ so that $\displaystyle du = 2x$

    Let $\displaystyle dv = -12x^3(4 - 3x^4)^{\frac{1}{3}}$ so that $\displaystyle v = \frac{3}{4}(4 - 3x^4)^{\frac{4}{3}}$ and the integral becomes


    $\displaystyle -\frac{1}{12}\int{x^2\left(-12x^3\right)(4 - 3x^4)^{\frac{1}{3}}\,dx} = -\frac{1}{12}\left[\frac{3}{4}x^2(4 - 3x^4)^{\frac{4}{3}} - \int{\frac{3}{2}x(4 - 3x^4)^{\frac{4}{3}}\,dx}\right]$

    $\displaystyle = -\frac{1}{16}x^2(4 - 3x^4)^{\frac{4}{3}} + \frac{1}{8}\int{x(4 - 3x^4)^{\frac{4}{3}}\,dx}$.


    Now apply integration by parts again.
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  4. #4
    Junior Member erich22's Avatar
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    Sir Prove it, I apply the integration by parts again, but I can get the answer. Could u help me end this problem?
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