Results 1 to 5 of 5

Math Help - Integral with trig-substitution

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    196
    Thanks
    1

    Integral with trig-substitution

    So I'm pretty much stuck with an integral problem. It was on the form (1 + (2t)^2)^0,5, using trig substitution I got to the integral 1/2(sec x)^3. Continuing with integration by parts, I finally got the solution (INT)1/2(sec x)^3 = 1/4(sec x)(tan x) + 1/4 ln |sec x + tan x| which should be correct. However, when I try to substitute back, using t = 1/2 tan x, my answer does not equal the answer in my book. And I can't seem to figure out what I did wrong... I've uploaded a picture of the last part of the problem, with the answer in my book below my answer.

    Any help would be greatly appreciated!
    Attached Thumbnails Attached Thumbnails Integral with trig-substitution-img702.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    773

    Re: Integral with trig-substitution

    Hello, gralla55!

    Your book's answer is correct, but it's rather clumsy.


    \int \sqrt{1 + (2t)^2}\,dt

    I would avoid those silly fractions . . .

    \text{Let }2t \:=\:\tan\theta \quad\Rightarrow\quad t \:=\:\tfrac{1}{2}\tan\theta \quad\Rightarrow\quad dt \:=\:\tfrac{1}{2}\sec^2\!\theta\,d\theta
    And: . \sqrt{1+(2t)^2} \:=\:\sqrt{1+\tan^2\!\theta} \:=\:\sqrt{\sec^2\!\theta} \:=\:\sec\theta

    Substitute: . \int \sec\theta\left(\tfrac{1}{2}\sec^2\!\theta\,d \theta \right) \;=\;\tfrac{1}{2}\int\sec^3\!\theta\,d\theta

    . . . . . . . . . =\;\tfrac{1}{4}\bigg(\sec\theta\tan\theta + \ln|\sec\theta + \tan\theta|\bigg) + C .[1]


    Back-substitute: . \tan\theta \,=\,\frac{2t}{1} \,=\,\frac{opp}{adj}

    \theta is in a right triangle with: . opp = 2t,\;adj = 1
    Hence: . hyp \,=\,\sqrt{1+4t^2}
    And: . \sec\theta \,=\,\frac{hyp}{adj} \,=\,\frac{\sqrt{1+4t^2}}{1} \,=\,\sqrt{1+4t^2}

    Substitute into [1]: . \tfrac{1}{4}\bigg(\sqrt{1+4t^2}\cdot 2t + \ln\left|\sqrt{1+4t^2} + 2t\right|\bigg) + C

    . . . . . . . . . . . . . =\;\tfrac{1}{2}\:\!t\sqrt{1+4t^2} + \tfrac{1}{4}\ln\left|2t + \sqrt{1+4t^2}\right| + C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    196
    Thanks
    1

    Re: Integral with trig-substitution

    Hi and thank you for very much for your help! I see you got the exact same answer I got, except I factored in the (1/2) under the first square root sign. What I can't seem to understand, is how the book gets from ln(2t + SQRT(1 + 4t^2)) to ln(t + SQRT((1/4 + t^2)). How are these two things the same??

    Thanks again!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,116
    Thanks
    1000

    Re: Integral with trig-substitution

    they're not "the same" ...

    \ln(2t + \sqrt{1 + 4t^2}) =

    \ln \left[2t + \sqrt{4\left(\frac{1}{4} + t^2\right)}\right] =

    \ln \left[2t + 2\sqrt{\frac{1}{4} + t^2}\right] =

    \ln \left[2\left(t + \sqrt{\frac{1}{4} + t^2}\right)\right] =

    \ln{2} + \ln\left(t + \sqrt{\frac{1}{4} + t^2}\right)

    ... but they do differ by a constant.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    196
    Thanks
    1

    Re: Integral with trig-substitution

    How did I not think of that.... Thank you very much! Now I can finally go to sleep, haha.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integral - trig substitution?
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 21st 2011, 05:49 PM
  2. Integral using Trig Substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 21st 2011, 07:25 AM
  3. Trig substitution integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 26th 2010, 10:20 PM
  4. Integral- trig substitution?
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 23rd 2010, 03:49 PM
  5. Trig Substitution Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 18th 2008, 07:16 AM

Search Tags


/mathhelpforum @mathhelpforum