Results 1 to 10 of 10

Math Help - Help with integration

  1. #1
    Newbie
    Joined
    Aug 2009
    Posts
    7

    Help with integration

    here are 2 sample questions i need help finding out how to solve:
    1.) ∫ x(sin^-1(x))dx

    2.) ∫ [(x+1)/√(4-(x^2))]dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by zachattack View Post
    here are 2 sample questions i need help finding out how to solve:
    1.) ∫ x(sin^-1(x))dx

    2.) ∫ [(x+1)/√(4-(x^2))]dx
    For the first one, use integration by parts, with u=\sin^{-1}x and \,dv=x\,dx.

    For the second one, note that

    \int\frac{x+1}{\sqrt{4-x^2}}\,dx=\int\frac{x\,dx}{\sqrt{4-x^2}}+\int\frac{\,dx}{\sqrt{4-x^2}}.

    The first of these two integrals requires a substitution, and the second integral should be a familiar one to you.

    Can you take it from here? If you still need assistance, please post back.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    7
    for the second problem, would trigonometric substitution be used then?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by zachattack View Post
    for the second problem, would trigonometric substitution be used then?
    It could...but that was a little more than what I was thinking.

    You can use the fact that \int\frac{\,dx}{\sqrt{1-x^2}}=\sin^{-1}x+C. So manipulate the second integral such that you can use this result.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    You need to apply integration by parts.

     \int vu' = uv - \int v'u

    In your first equation make

    v = x

    u'=\sin^{-1}x
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2009
    Posts
    7
    ok I get how the second integral is arcsin, but I don't know how to manipulate the first integral
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,234
    Thanks
    27
    Quote Originally Posted by pickslides View Post
    You need to apply integration by parts.

     \int vu' = uv - \int v'u

    In your first equation make

    v = x

    u'=\sin^{-1}x
    from here

    find v' and u

    then use \int vu' = uv - \int v'u
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2009
    Posts
    7
    Is there anyway someone could just do both problems and post the work?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by zachattack View Post
    Is there anyway someone could just do both problems and post the work?
    no just use the hints given.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zachattack View Post
    Is there anyway someone could just do both problems and post the work?
    MHF does not typically provide complete solutions. Effort on your part is expected. Please make an attempt based on the help you've been given. Further help, if required, based on where you get stuck or make a mistake can then be given.
    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