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Math Help - I need some helps in integrations !!! Really URGENT

  1. #1
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    I need some helps in integrations !!! Really URGENT

    integrate:

    1. 1/ x^2* sqr(x^2+4) dx sqr is square root

    2. e^(2x)*sin(e^x)dx ( I know I should do by parts but I am stucked)

    3. 1/ sqr(x)*(x^(1/3)+1)dx

    4. x^(2)+1 / (x^(3)-4x^2)dx

    5. 1/ (x* sqr(9+4*x^(2)))dx

    6.x^(2)/ sqr(2x-x^(2)) dx

    7. cost / (2 -(sint)^(2)) dt ( I guess it's an exact derivative here but I can't move on)

    8. x^(3)/ (a^(2)-x^2) dx I can't move on with trig substitution

    9. lnx / (x+1)^2 dx

    10. e^(ax) * cos(bx) dx By Parts ? how can I do it ?

    11. x* e^(sqr(x))dx

    12. Arctan (sqr(x)) dx

    13. (tan(x))^6 dx

    14. (sec(x))^6dx

    15. x^(1/3) / (1+x) dx

    Some more to come !!

    Please give me some help. I really need it.
    Last edited by CaptainBlack; October 31st 2008 at 12:04 AM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    integrate:

    1. 1/ x^2* sqr(x^2+4) dx sqr is square root

    2. e^(2x)*sin(e^x)dx ( I know I should do by parts but I am stucked)

    3. 1/ sqr(x)*(x^(1/3)+1)dx

    4. x^(2)+1 / (x^(3)-4x^2)dx

    Some more to come !!

    Please give me some help. I really need it.
    1. Let x=\frac{1}{u}
    2. Let e^x=u and realize that e^{2x}=e^x\cdot{e^x}

    3. Let x=u^6

    4.Partial fraction decomp.
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  3. #3
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    Can you explain more detail please ?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    Can you explain more detail please ?
    Which ones in particular are you having difficulties with?
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  5. #5
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    All actually, but can we start with the second one first ?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    All actually, but can we start with the second one first ?
    \int{e^{2x}\sin(e^x)}dx=\int{e^x\cdot{e^x}\sin\lef  t(e^x\right)dx}

    If we let u=e^x\Rightarrow{du=e^xdx}

    So out integral becomes

    \int{e^u\sin(u)}du

    Which can be easily computed with two iterations of integration by parts
    Last edited by Mathstud28; October 30th 2008 at 06:41 PM. Reason: forgot du
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  7. #7
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    I thought e^2x = u^2 ?

    how about the others ?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    integrate:



    5. 1/ (x* sqr(9+4*x^(2)))dx

    6.x^(2)/ sqr(2x-x^(2)) dx

    7. cost / (2 -(sint)^(2)) dt ( I guess it's an exact derivative here but I can't move on)

    8. x^(3)/ (a^(2)-x^2) dx I can't move on with trig substitution

    9. lnx / (x+1)^2 dx

    10. e^(ax) * cos(bx) dx By Parts ? how can I do it ?

    Some more to come !!

    Please give me some help. I really need it.
    5. Let 2x=3\tan(\theta)

    6. Complete the square on the bottom and let x+1=u

    7. Let u=\sin(x)\Rightarrow{du=\cos(x)dx}

    8. Use polynomial divison first

    9. Integrate by parts...you could \ln(x)=u first if you were so inclined

    10. yes parts...just remember that the a and b are constants
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  9. #9
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    7. That's how I started but the n I got in the problem of the integrating 1/2-u^(2)
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    7. That's how I started but the n I got in the problem of the integrating 1/2-u^(2)
    2-u^2=\left(\sqrt{2}-u\right)\left(\sqrt{2}+u\right)

    Use P-fracs
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  11. #11
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    Do you mind explain a little bit more detail about each one ?

    I am kinda weak in P fraction so How can I decompose number 4 ?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by nns91 View Post
    Do you mind explain a little bit more detail about each one ?

    I am kinda weak in P fraction so How can I decompose number 4 ?
    Sorry but it is way past my bedtime...If no one has finished these I will do so tomorrow.
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  13. #13
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    Hello, nns91!

    1)\;\;\int \frac{dx}{x^2\sqrt{x^2+4}}
    Let x = 2\tan\theta\quad\Rightarrow\quad dx = 2\sec^1\theta\,d\theta

    Substitute: . \int\frac{2\sec^2\!\theta\,d\theta}{4\tan^2\!\thet  a\cdot2\sec\theta} \;=\;\tfrac{1}{4}\int \frac{\sec\theta}{\tan^2\theta}\,d\theta . . . etc.




    2)\;\; e^{2x}\sin(e^x)\,dx
    \begin{array}{ccccccc}u &=& e^x & & dv &=&e^x\sin(e^x)\,dx \\ du &=& e^x\,dx & & v &=& \text{-}\cos(e^x) \end{array}

    We have: . \text{-}e^x\cos(e^x) + \int e^x\cos(e^x)\,dx \;=\;\text{-}e^x\cos(e^x) + \sin(e^x) + C




    3)\;\; \int \frac{dx}{\sqrt{x}(x^{\frac{1}{3}}+1)}
    Let u = x^{\frac{1}{6}} \quad\Rightarrow\quad x = u^6 \quad\Rightarrow\quad dx = 6x^5\,du
    . . Also: . x^{\frac{1}{2}} = u^3,\;\;x^{\frac{1}{3}} = u^2

    Substitute: . \int\frac{6u^5\,du}{u^3(u^2+1)} \;=\;6\!\!\int\frac{u^2}{u^2+1}\,du . . . etc.




    4)\;\;\int\frac{x^2+1}{x^3-4x^2}\,dx
    Partial Fractions: . \frac{x^2+1}{x^2(x-4)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-4}

    . . and we get: . A = \text{-}\tfrac{1}{16},\;B = \text{-}\tfrac{1}{4},\;C = \tfrac{17}{16}

    So we have: . \int\left[\frac{\text{-}\frac{1}{16}}{x} + \frac{\text{-}\frac{1}{4}}{x^2} + \frac{\frac{17}{16}}{x-4}\right]\,dx \;=\;\text{-}\tfrac{1}{16}\!\!\int\frac{dx}{x} - \tfrac{1}{4}\!\!\int x^{-2}dx + \tfrac{17}{16}\!\!\int\frac{dx}{x-4} \quad\hdots

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  14. #14
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    For Qn 13 and 14. You can refer to the link below for step by step explanation and guide. Hope it helps.

    Antiderivatives / Reduction Formulas - 2
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