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Math Help - Integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Integral

    I have a portion of ad admittance exam coming up tomorrow and I was wondering if someone could give me some integrals I could solve and then you could check my work...I am getting fairly good at integrals but practice never hurt! Now I dont want ridculously difficult integrals...but I don't want too easy of ones either xD

    If someoen could give me some I would be very greatful

    Mathstud
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  2. #2
    Moo
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    Hello,

    Try this one :

    \int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt

    There's nothing great about this, but take care of careless mistakes
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  3. #3
    o_O
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    A couple I found from the MIT integration bee from a while back:

    \int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx

    \int \sin(101x) \cdot \sin^{99} (x) dx

    \int e^{\arccos (x)} dx

    Have fun?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    Try this one :

    \int_n^{n+1} \arctan \left(\frac{1}{\sqrt{t}} \right) dt

    There's nothing great about this, but take care of careless mistakes
    Thanks a lot Moo! I appreciate it...this is a fairly easy one...or at least I hope xD

    \int{arctan\bigg(\frac{1}{\sqrt{x}}\bigg)dx}=\int{  arcot(\sqrt{x})dx}

    Then you would go
    Let u=\sqrt{x}\Rightarrow{u^2=x}
    Then dx=2udu
    So we have \int{2u\cdot{arcot(u)du}}

    Using parts we have \int{2u\cdot{arcot(u)}}du=u^2\cdot{arcot(u)}+\int\  frac{u^2}{1+u^2}

    THis is the same as
    u^2\cdot{arcot(u)}+\bigg[\int\frac{u^2+1}{u^2+1}du-\int\frac{1}{u^2+1}du\bigg]

    Which then obviously goes to

    u^2\cdot{arcot(u)}+u-arctan(u)

    Back subbing we get x\cdot{arcot(\sqrt{x})}+\sqrt{x}-arctan(\sqrt{x})

    Whew I dont know why that took so long to write ...I showed all my steps just so you could critique me. Thanks
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  5. #5
    Moo
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    Actually, I've never used cot function...
    Yes, the answer looks like yours

    Try this one (I don't have the answer yet) :

    \int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    A couple I found from the MIT integration bee from a while back:

    \int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx

    \int \sin(101x) \cdot \sin^{99} (x) dx

    \int e^{\arccos (x)} dx

    Have fun?
    Ok well lets do these one at a time...starting with number three

    \int{e^{arccos(x)}dx}

    Letting u=arcos(x)\Rightarrow{cos(u)=x}
    Then dx=-sin(u)
    So we have \int-\sin(u)e^udu
    then by parts we haev -sin(u)e^{u}+\int\cos(u)e^{u}du
    which then goes to -sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}

    Adding the like integrals and dividing by 2 we get \int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))


    back subbing we get \int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})
    Last edited by Mathstud28; April 28th 2008 at 03:00 PM.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Actually, I've never used cot function...
    Yes, the answer looks like yours

    Try this one (I don't have the answer yet) :

    \int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx
    Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series
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  8. #8
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    Aww moo...Do I really have to do that one...you would have to use the binomial coefficient power series
    You said you wanted to train
    I doubt your test will be about Krizalid-like integrals
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    You said you wanted to train
    I doubt your test will be about Krizalid-like integrals
    lasjflsdfkjasldfkjsadf;ljk......ok I will do it then
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    A couple I found from the MIT integration bee from a while back:

    \int \left(\frac{1}{\ln x} + \ln \left(\ln x\right)\right)dx

    \int \sin(101x) \cdot \sin^{99} (x) dx

    \int e^{\arccos (x)} dx

    Have fun?
    For the third one we have \int\frac{1}{\ln(x)}-\ln(\ln(x))dx
    Letting u=\ln(x)\Rightarrow{e^u=x}
    then dx=e^{u}
    so we have \int{\frac{e^{u}}{u}+e^{u}\ln(u)du}

    which by splitting the integrals gives \int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy

    leaving the first as it is we do integration by parts on the second to reveal

    \int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u  }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]

    cancelling like terms we get \int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)

    Back subbing we get \int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    Actually, I've never used cot function...
    Yes, the answer looks like yours

    Try this one (I don't have the answer yet) :

    \int_0^1 \sqrt[3]{1-x^7}-\sqrt[7]{1-x^3} dx
    e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{3}}dx-e\int_0^{1}\bigg[\ln\bigg(\frac{1}{x}\bigg)\bigg]^{\frac{1}{7}}dx?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok well lets do these one at a time...starting with number three

    \int{e^{arccos(x)}dx}

    Letting u=arcos(x)\Rightarrow{cos(u)=x}
    Then dx=-sin(u)
    So we have \int-\sin(u)e^udu
    then by parts we haev -sin(u)e^{u}+\int\cos(u)e^{u}du
    which then goes to -sin(u)e^{u}+\cos(u)e^{u}-\int{-\sin(u)e^{u}du}

    Adding the like integrals and dividing by 2 we get \int-\sin(u)e^{u}du=\frac{e^{u}}{2}(\cos(u)-\sin(u))


    back subbing we get \int{e^{arccos(x)}dx}=\frac{e^{arccos(x)}}{2}(x-\sqrt{1-x^2})
    Quote Originally Posted by Mathstud28 View Post
    For the third one we have \int\frac{1}{\ln(x)}-\ln(\ln(x))dx
    Letting u=\ln(x)\Rightarrow{e^u=x}
    then dx=e^{u}
    so we have \int{\frac{e^{u}}{u}+e^{u}\ln(u)du}

    which by splitting the integrals gives \int{\frac{e^{u}}{u}du}+\int{e^{u}\ln(u)}dy

    leaving the first as it is we do integration by parts on the second to reveal

    \int{\frac{e^{u}}{u}+e^{u}\ln(u)}du=\int\frac{e^{u  }}{u}du+\bigg[e^{u}\ln(u)-\int\frac{e^{u}}{u}du\bigg]

    cancelling like terms we get \int{\frac{e^u}{u}+e^{u}\ln(u)}du=e^{u}\ln(u)

    Back subbing we get \int\bigg[\frac{1}{\ln(x)}+\ln(\ln(x))dx\bigg]=e^{\ln(x)}\cdot{\ln(\ln(x))}=x\ln(\ln(x))+C
    Are these right o_O?
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  13. #13
    o_O
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    Looks good to me. You made some typos but that's probably due to messing up while typing up your work.

    How about this one:
    \int_{0}^{\infty} xe^{-x^{3}}dx
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by o_O View Post
    Looks good to me. You made some typos but that's probably due to messing up while typing up your work.

    How about this one:
    \int_{0}^{\infty} xe^{-x^{3}}dx
    \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{3n+2}}{(3n+2)n!}\bigg|_{0}^{\infty}

    Is that right?
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  15. #15
    Math Engineering Student
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    Do you know when you have to apply power series for an integral? Did you think first to make a substitution?
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