Results 1 to 8 of 8

Math Help - Help with Integration

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    38

    Help with Integration

    Integrate:
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by lacy1104 View Post
    Integrate:
    This does not have an antiderivative
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2008
    Posts
    38
    This is what the problem really says. I thought by doing that integral was how you did it.

    Show that for on [-1,1], the expression for aj is given by:
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by lacy1104 View Post
    This is what the problem really says. I thought by doing that integral was how you did it.

    Show that for on [-1,1], the expression for aj is given by:
    and what is a_j exactly?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2008
    Posts
    38
    Quote Originally Posted by Jhevon View Post
    and what is a_j exactly?

    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by lacy1104 View Post
    two more discrepancies here. in the first post you had \sqrt{1 ~{\color{red}+}~ x^2} in the denominator, which made the integral impossible.

    secondly, now this T_i(x) appears and you never told us anything about it. Please try to describe the problem completely, we do not have your text book
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Apr 2008
    Posts
    38
    Quote Originally Posted by lacy1104 View Post
    This is what the problem really says. I thought by doing that integral was how you did it.

    Show that for on [-1,1], the expression for aj is given by:
    I was wrong in the first entry. it is -

    T0(x) = 1, T1(x)=x
    In general, Ti(x)= cos[i arccos x], i > 0

    It is posted on this website (its question 3b)
    http://www.math.niu.edu/~dattab/math435/HW4.pdf
    Follow Math Help Forum on Facebook and Google+

  8. #8
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by lacy1104 View Post
    I was wrong in the first entry. it is -

    T0(x) = 1, T1(x)=x
    In general, Ti(x)= cos[i arccos x], i > 0

    It is posted on this website (its question 3b)
    http://www.math.niu.edu/~dattab/math435/HW4.pdf
    if it should have been minus, then the first integral is no problem. make a substitution of u = \arccos x, you will get an integral that you can do integration by parts on
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

Search Tags


/mathhelpforum @mathhelpforum