Results 1 to 4 of 4

Math Help - trig substitutions

  1. #1
    Junior Member
    Joined
    Jul 2007
    Posts
    72

    trig substitutions

    I have having trouble doing this one:
    integral of radical(x^2 + 4)dx
    I set x = tan(u)
    dx = sec^2(x)du
    here is what i did:
    integral of radical(4tan^2(u) + 4)
    Pulled out a radical 4
    integral of 2tan(u)
    this equals 2sec^2(u) + C
    sec(u) in this scenario is (x+2)/(2) since tan(u) = x/2
    I did this then:
    2((x+2/2))^2
    The answer after doing this isnt the same in the book
    I am suppose to get: 1/2xradical(x^2 + 4) + 2ln(x + radical(x^2 + 4)) + C
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    \int\sqrt{x^{2}+4}dx

    Let x=2tan(u), \;\ dx=2sec^{2}(u)du

    \int\sqrt{(2tan(u))^{2}+4}\cdot{2sec^{2}(u)}du

    \int\sqrt{4tan^{2}(u)+4}\cdot{2sec^{2}(u)}du

    \int\sqrt{4(tan^{2}(u)+1)}\cdot{2sec^{2}(u)}du

    \int(2sec(u))2sec^{2}(u)du

    4\int{sec^{3}(u)}du

    \int{sec^{n}(u)}du=\frac{sec^{n-2}(u)tan(u)}{n-1}+\frac{n-2}{n-1}\int{sec^{n-2}(u)}du

    After you evaluate the integral, you can resub u=tan^{-1}(\frac{x}{2}) to get it back in terms of x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by davecs77 View Post
    I have having trouble doing this one:
    integral of radical(x^2 + 4)dx
    I set x = tan(u)
    dx = sec^2(x)du
    here is what i did:
    integral of radical(4tan^2(u) + 4)
    Pulled out a radical 4
    integral of 2tan(u)
    this equals 2sec^2(u) + C
    sec(u) in this scenario is (x+2)/(2) since tan(u) = x/2
    I did this then:
    2((x+2/2))^2
    The answer after doing this isnt the same in the book
    I am suppose to get: 1/2xradical(x^2 + 4) + 2ln(x + radical(x^2 + 4)) + C
    The number in red above is an error. And I more or less can't follow what you did after that. So....
    \int \sqrt{x^2 + 4}dx

    Use x = 2tan(u) --> dx = 2sec^2(u) du

    So
    \int \sqrt{x^2 + 4}dx = \int \sqrt{4tan^2(u) + 4} \cdot 2 sec^2(u) du

    = 2 \int \sqrt{4(tan^2(u) + 1)}sec^2(u) du

    = 4 \int \sqrt{tan^2(u) + 1}sec^2(u) du

    = 4 \int \sqrt{sec^2(u)}sec^2(u) du

    = 4 \int sec(u) \cdot sec^2(u) du

    = 4 \int sec^3(u) du

    Can you do it from here?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Also

    \int\sec^3x~dx=\color{red}\int\sec{x}\sec^2x~dx

    Then you can apply integration by parts, not hard.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig substitutions
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 21st 2008, 10:46 PM
  2. Trig Substitutions
    Posted in the Calculus Forum
    Replies: 15
    Last Post: April 21st 2008, 02:37 PM
  3. Trig Substitutions Need Help
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 4th 2007, 10:13 AM
  4. trig substitutions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 1st 2007, 04:15 AM
  5. trig substitutions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 29th 2007, 01:33 PM

Search Tags


/mathhelpforum @mathhelpforum