Results 1 to 11 of 11

Math Help - Integration

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    105

    Integration

    I hate it. Stuck on 2 integration problems

    a). \int\frac{1}{\sqrt{x}}\sin\sqrt{x}\ .dx\ where\ x = u^2

    b). \int\frac{2}{e^{2x} + 4}\ where\ u = e^{2x} + 4

    Thanks for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Quote Originally Posted by Beard View Post
    I hate it. Stuck on 2 integration problems

    a). \int\frac{1}{\sqrt{x}}\sin\sqrt{x}\ .dx\ where\ x = u^2

    b). \int\frac{2}{e^{2x} + 4}\ where\ u = e^{2x} + 4

    Thanks for any help
    Your first substitution should work giving

    2\int \sin u\, du.

    For the second try

    u = e^{x}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2008
    Posts
    792
    For the second, I prefer u = e^{-x}. No partial fractions decomposition required, just simple sub.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,347
    Thanks
    30
    Quote Originally Posted by Chop Suey View Post
    For the second, I prefer u = e^{-x}. No partial fractions decomposition required, just simple sub.
    Definitely quicker (as a little easier) but not obvious for most.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2008
    Posts
    105
    Thanks for the first part, however I am unable to do the second part of your reply where you said try:

    u = e^x

    because the book has given me a 'set' substitution so it must remain as

    u = e^{2x} + 4
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Beard View Post
    Thanks for the first part, however I am unable to do the second part of your reply where you said try:

    u = e^x

    because the book has given me a set 'substitution' so it must remain as

    u = e^{2x} + 4
    In that case then, if u=e^{2x}+4, then \,du=2e^{2x}\,dx

    Thus, the integral becomes \int\frac{2\,dx}{u}=\int\frac{2e^{2x}\,dx}{e^{2x}u  }

    Since u=e^{2x}+4, that means that e^{2x}=u-4

    So the integral becomes \int\frac{\,du}{u\left(u-4\right)}

    And now, you'll need to apply the technique of partial fractions.
    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
    13
    Quote Originally Posted by Chris L T521 View Post

    you'll need to apply the technique of partial fractions.
    No, haha, because of u-(u-4)=4.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2008
    Posts
    105
    Thus, the integral becomes

    Confused by the second part of this unless you are saying that you are multiplying by 1 in the form of \frac{e^{2x}}{e^{2x}}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Beard View Post
    Thus, the integral becomes

    Confused by the second part of this unless you are saying that you are multiplying by 1 in the form of \frac{e^{2x}}{e^{2x}}
    Yes. I multiplied by 1 in the form of \frac{e^{2x}}{e^{2x}} because now we can replace 2e^{2x}\,dx with \,du in the numerator. In the denominator, you can now replace e^{2x} with u-4.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2008
    Posts
    105
    So how do you integrate

    \frac{du}{u^2 - 4u}

    ?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    \int\frac{1}{u(u-4)}du

    One way is split into partial fractions.

    \frac{1}{4}\int\frac{1}{u-4}du-\frac{1}{4}\int\frac{1}{u}du

    Now, it's easy to see each one is in terms of ln.
    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