Results 1 to 9 of 9

Math Help - Trig Substitution

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    26

    Trig Substitution

    The problem is the integral of (x^3)/(((x^2)+16)^.5) dx x=4tan(theta)
    I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta, but don't know what to do with it. Any help is appreciated.

    Thanks
    Last edited by Hellacious D; September 29th 2009 at 02:49 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    \frac{x^3}{x^2 + 16} = \frac{x(x^2 + 16) - 16x}{x^2 + 16} = (split the numerator)

    (The give-away is the higher power top than bottom - or does it have to be trig?)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    26
    My apologies, I have been up all night and am feeling stupid. it's actually the square root of x^2 + 16 on bottom which is what requires the trig.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Still the x^3 on top means you don't need trig... use - Wolfram|Alpha and click on show steps, but I'll post a picture, just in case it helps
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Quote Originally Posted by Hellacious D View Post
    The problem is the integral of (x^3)/((x^2)+16) dx x=4tan(theta)
    I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta, but don't know what to do with it. Any help is appreciated.

    Thanks
    Ok so you can write the integral as this:

    I(x)=\frac{x^3}{x^2+16}=\frac{x^3+16x-16x}{x^2+16}=\frac{x(x^2+16)}{x^2+16}-\frac{16x}{x^2+16}= x-\frac{16x}{x^2+16}
    <br />
\int{I(x)} = \int {xdx}-\int {\frac{16x}{x^2+16}}

    Put t = x^2+16
    So,

    dt = 2xdx and 8dt = 16xdx

    Substitute all this,

    I=\frac{x^2}{2}-\int{\frac{8dt}{t}}
    which gives
    I=\frac{x^2}{2}-8ln|t|+c

    back substitute t to get

    I=\frac{x^2}{2}-8ln|x^2+16|+c

    which is your answer, no trigonometry required!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Sep 2009
    Posts
    26
    Tom, I appreciate your pointing out of that tool for checking answers, but the way in which it gets answers is nothing like how I'm being taught and doesn't appear to be something I can replicate. tanujkush, sorry for wasting your time. I realized I had put x^2 + 16 and meant to put ((x^2)+16)^.5 on bottom. It has been fixed now.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Quote Originally Posted by Hellacious D View Post
    I've worked it down to 4^2 times the integral of (tan^3(theta))/(sec(theta)) dtheta
    Whoops... multiply this by the derivative of 4 tan theta, though. Pic coming
    Last edited by tom@ballooncalculus; September 29th 2009 at 03:30 AM. Reason: sorry, theta not x
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Sep 2009
    Posts
    13
    Quote Originally Posted by Hellacious D View Post
    Tom, I appreciate your pointing out of that tool for checking answers, but the way in which it gets answers is nothing like how I'm being taught and doesn't appear to be something I can replicate. tanujkush, sorry for wasting your time. I realized I had put x^2 + 16 and meant to put ((x^2)+16)^.5 on bottom. It has been fixed now.

    Oh then its even more simple,

    so your integrand is this i hope?

    I = \int {\frac{x^3}{\sqrt{x^2+16}}}dx

    Plug in t^2 = x^2+16 to get
    2tdt=2xdx or
    tdt=xdx

    This changes your integrand to

    I=\int {\frac{(t^2-16)tdt}{t}}

    which reduces, quite simply, to

    I=\int {(t^2-16)dt}

    I=t^3/3-16t+c

    back substitute t^2

    Answer=(\sqrt{(x^2+16)})^3-16\sqrt{x^2+16}+c
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,034
    Thanks
    49
    Should also work with trig, but since you're putting another function inside x you need to multiply what you had by the derivative of the inner function, according to the chain rule, which I like to depict so...



    Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (which is the inner function of the composite and hence subject to the chain rule).

    So, the problem so far...



    The chain rule also the key to the new integrand ...





    ________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus Forum

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; September 29th 2009 at 05:57 AM. Reason: ahem, 'typo'
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig substitution
    Posted in the Calculus Forum
    Replies: 15
    Last Post: June 29th 2010, 12:32 PM
  2. Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 23rd 2010, 01:12 PM
  3. Trig Substitution again
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 29th 2009, 11:01 AM
  4. Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 25th 2009, 06:11 AM
  5. Trig substitution
    Posted in the Calculus Forum
    Replies: 27
    Last Post: June 16th 2008, 06:55 PM

Search Tags


/mathhelpforum @mathhelpforum