Results 1 to 4 of 4

Math Help - Integral.

  1. #1
    Ted
    Ted is offline
    Member
    Joined
    Feb 2010
    From
    China
    Posts
    199
    Thanks
    1

    Integral.

    Hello.
    evaluate:

    \int \sqrt{x^2+16} dx.
    Without using Trig/Hyperbolic Substitutions.
    And you must use the substitution method only.
    I believed, its impossible to solve it.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2009
    Posts
    64
    The way I was taught to answer integrals:
    \sqrt{x^2+16} dx

     <br />
=(x^2+16)^\frac{1}{2}dx<br />

     <br />
u = x^2+16 =><br />
du = 2x<br />

     <br />
=\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]<br />
     <br />
=\frac{1}{2}*<br />
\frac{u^{3/2}}{\frac{3}{2}}<br />

    Answer:
     <br />
= \frac{1}{3}(x^2+16)^{3/2}<br />
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,977
    Thanks
    1643
    Quote Originally Posted by lil_cookie View Post
    The way I was taught to answer integrals:
    \sqrt{x^2+16} dx

     <br />
=(x^2+16)^\frac{1}{2}dx<br />

     <br />
u = x^2+16 =><br />
du = 2x<br />

     <br />
=\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]<br />
    Unfortunately, you cannot just put that "x" into the integrand!

     <br />
=\frac{1}{2}*<br />
\frac{u^{3/2}}{\frac{3}{2}}<br />

    Answer:
     <br />
= \frac{1}{3}(x^2+16)^{3/2}<br />
    No, the derivative of \frac{1}{3}(x^2+ 16)^{3/2} is
    (3/2)(1/3)(x^2+ 16)^{1/2} (2x)= x(x^2+16)^{1/2}, not just (x^2+ 16)^{1/2}

    I doubt that you were taught that way!

    Ted, I don't believe there is any way to do that other than either the trig or hyperbolic substitutions.

    Of course, you could always look it up in a table of integrals. Does that count?
    http://www.integral-table.com/
    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
    Quote Originally Posted by Ted View Post
    Hello.
    evaluate:

    \int \sqrt{x^2+16} dx.
    Without using Trig/Hyperbolic Substitutions.
    put x=4u and the integral becomes 16\int\sqrt{u^2+1}\,du.

    put t=\sqrt{u^2+1}+u then \frac1t=\sqrt{u^2+1}-u, thus \frac12\left(t+\frac1t\right)=\sqrt{u^2+1} and \frac12\left(t-\frac1t\right)=u so i think you can finish it.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum