Results 1 to 5 of 5

Math Help - Trigonometric Substitution Integral

  1. #1
    Del
    Del is offline
    Member
    Joined
    Nov 2007
    Posts
    83

    Trigonometric Substitution Integral

    Evaluate the indefinite integral:


    ⌡ [√(16 x^2 − 144) / x] dx

    Sorry for the lame text, but it's all I could gather, I've gotten to the point:

    12 ⌠
    ⌡ [cos (t) / cos^3 (t) sin (t)] dt

    by setting x = 3 tan (t); dx = 3 sec^2 (t); and √(x^2 + 9) = 3 sec (t), but I don't know how to solve it from here. Please help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by Del View Post
    Evaluate the indefinite integral:


    ⌡ [√(16 x^2 − 144) / x] dx
    You don't need to apply a trig. sub.

    Why don't you flip it around by settin' u^2=16x^2-144?

    As for mathematical symbols, see my signature.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2008
    Posts
    91
    Quote Originally Posted by Krizalid View Post
    You don't need to apply a trig. sub.

    Why don't you flip it around by settin' u^2=16x^2-144?

    As for mathematical symbols, see my signature.
    But the x is on the bottom in his equation so by your substitution you aren't going to eliminate x you are going to make it x^2 on the bottom.
    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
    And by settin' that substitution x^2  = \frac{{u^2  + 144}}<br />
{{16}}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    610
    Hello, Del!

    Integrate: . \int\frac{\sqrt{16x^2-144}}{x}\,dx

    Factor out 16: . 4\int\frac{\sqrt{x^2-9}}{x}\,dx

    Let x \:= \:3\sec\theta\quad\Rightarrow\quad dx\:=\:3\sec\theta\tan\theta\,d\theta

    Substsitute: . 4\int\frac{3\tan\theta}{3\sec\theta}\,(3\sec\theta  \tan\theta\,d\theta) \;=\;12\int\tan^2\!\theta\,d\theta

    We have: . 12\int(\sec^2\!\theta - 1)\,d\theta \;=\;12(\tan\theta - \theta) + C

    Back-substitute: . 12\left(\frac{\sqrt{x^2-9}}{3} \,- \,\text{arcsec}\frac{x}{3}\right) + C

    Answer: . 4\left(\sqrt{x^2-9} \,- \,3\!\cdot\!\text{arcsec}\frac{x}{3}\right) + C

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Annoying trigonometric substitution integral
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 1st 2011, 03:50 PM
  2. [SOLVED] Very last step of a trigonometric integral substitution?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 12th 2011, 06:15 AM
  3. Integral - trigonometric substitution?
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 17th 2010, 05:58 PM
  4. Trigonometric Substitution Integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 7th 2010, 01:22 AM
  5. Replies: 5
    Last Post: July 24th 2009, 07:41 PM

Search Tags


/mathhelpforum @mathhelpforum