Results 1 to 12 of 12

Math Help - Integrals!

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    2

    Integrals!

    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2007
    From
    London / Cambridge
    Posts
    591
    for the second one do integration by parts, you should get an answer of

    \frac{x^2 lnx}{2} - \frac{x^2}{4}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Aug 2007
    From
    USA
    Posts
    3,111
    Thanks
    2
    The definition of "tough" always has eluded me.

    Frequent Definition: I don't know how to do it.

    Occasional Definition: I can do it, but it takes too long and requires too much effort.

    Pratical Definition: It can be done, but who would want to?

    Reasonable Definition: I could spend a lot of time on this, using some way that I have concocted, but even if it works, there simply MUST be an easier way.

    This leads me to ask two things:

    1) Why have you been given these integrals? Are you REALLY expected to solve them in a closed form? Without a very specific hint, and with your missing he Integration by Parts on the second, it seems to me that the first is not reasonable. Do you have a specific hint?

    2) Does one get to use numerical methods? Numbers are not evil. It's okay if that's what we have.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Mess View Post
    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
    For the first one, try substituting x^2 = tan(t).

    You will end up with an integrand looking like:

    \frac{\sqrt{sec(t)-tan(t)}}{sec^2(t)}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Here's a specific calculation in this case.

    Quote Originally Posted by Mess View Post

    Evaluate \int_0^1 {x\ln x\,dx}
    \ln x=\int_1^x\frac1u\,du. So,

    \int_0^1 {x\ln x\,dx} = - \int_0^1 {\int_x^1 {\frac{x}<br />
{u}\,du} \,dx} = - \int_0^1 {\underbrace {\int_0^u {x\,dx} }_{\frac{1}<br />
{2}u^2 }\frac{1}<br />
{u}\,du} .

    The rest is routine, and the answer is -\frac14.

    Even I'd like to take a more general case:

    Let m,n>0.

    Define \int_0^1x^m\ln^nx\,dx.

    Make substitution u=-\ln x which can be written as x=e^{-u}\implies dx=-e^{-u}\,du, the integral becomes

    \int_0^1 {x^m \ln ^n x\,dx} = ( - 1)^n \int_0^\infty {e^{ - (m + 1)u} u^n \,du} .

    Define another substitution according to

    \alpha = (m + 1)u \implies u = \frac{\alpha }<br />
{{m + 1}}\,\therefore \,du = \frac{1}<br />
{{m + 1}}\,d\alpha ,

    \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }}<br />
{{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}}<br />
{{(m + 1)^{n + 1} }}.

    --

    As for the first integral, set x=\frac1u.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Krizalid View Post
    \int_0^1 {x^m \ln ^n x\,dx} = \frac{{( - 1)^n }}<br />
{{(m + 1)^{n + 1} }}\int_0^\infty {e^{ - \alpha } \alpha ^n \,d\alpha } = \frac{{( - 1)^n n!}}<br />
{{(m + 1)^{n + 1} }}.
    I see you are begining to love the gamma function
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    Aha, yes! It's really cool!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    GAMMA Mathematics
    colby2152's Avatar
    Joined
    Nov 2007
    From
    Alexandria, VA
    Posts
    1,172
    Awards
    1
    Quote Originally Posted by Krizalid View Post
    Aha, yes! It's really cool!
    I have a love hate relationship with the Gamma function.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Mess View Post
    Hi people!

    I can't solve these ones, they're tough!

    Solve \int {\frac{{\sqrt {\sqrt {x^4 + 1} - x^2 } }}<br />
{{x^4 + 1}}\,dx}

    Evaluate \int_0^1 {x\ln x\,dx}

    Thanks!
    And the answer (which may influence your decision to persist in working it out) is .............. :
    Attached Thumbnails Attached Thumbnails Integrals!-integral-2.gif  
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    I wouldn't trust in the integrator, after substitution x=\frac1u leads another integral which can be solved easily.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Krizalid View Post
    I wouldn't trust in the integrator, after substitution x=\frac1u leads another integral which can be solved easily.
    The Integrator can be trusted, just not trusted to give the answer in its simplest form.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by topsquark View Post
    The Integrator can be trusted, just not trusted to give the answer in its simplest form.

    -Dan
    Mathematicians (the extremists) do not use calculator, computers, and machines.
    Because they shall say this is a direct violation against what math is about.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 10:23 PM
  2. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  3. Integrals : 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 08:40 AM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 05:52 PM
  5. integrals Help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2008, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum