Results 1 to 6 of 6

Math Help - Help with integral

  1. #1
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1

    Help with integral

    Can anyone do this integral for me? I am not convinced my answer is correct :/

    \int \sqrt{1+2e^{2t}}~dt

    Here is my work:

    ---------------------------------------------
    SUBSTITUTE:
    \frac{1}{\sqrt{2}}~tan(a)=e^t

    \frac{1}{\sqrt{2}}~sec^2(a)~da=e^t~dt

    \frac{1}{\sqrt{2}e^t}~sec^2(a)~da=dt

    cot(a)~sec^2(a)~da=dt

    csc(a)~sec(a)~da=dt
    ---------------------------------------------

    = \int csc(a)~sec(a) \sqrt{1+2\left(\frac{tan(a)}{\sqrt{2}}\right)^2}~d  a

    = \int csc(a)~sec(a) \sqrt{1+tan^2(a)}~da

    = \int csc(a)~sec^2(a)~da

    = \int csc(a)~(1+tan^2(a))~da

    = \int csc(a)~da + \int \frac{sin(a)}{cos^2(a)}~da

    = -ln| csc(a)+cot(a) | + sec(a)~da

    = -ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\rig  ht|+(2e^{2t}+1)^{-1/2} + C

    I've done it twice and gotten this answer, but it just seems kind of ugly and unlikely, and my calculator told me a different answer
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by angel.white View Post
    Can anyone do this integral for me? I am not convinced my answer is correct :/

    \int \sqrt{1+2e^{2t}}~dt

    Here is my work:

    ---------------------------------------------
    SUBSTITUTE:
    \frac{1}{\sqrt{2}}~tan(a)=e^t

    \frac{1}{\sqrt{2}}~sec^2(a)~da=e^t~dt

    \frac{1}{\sqrt{2}e^t}~sec^2(a)~da=dt

    cot(a)~sec^2(a)~da=dt

    csc(a)~sec(a)~da=dt
    ---------------------------------------------

    = \int csc(a)~sec(a) \sqrt{1+2\left(\frac{tan(a)}{\sqrt{2}}\right)^2}~d  a

    = \int csc(a)~sec(a) \sqrt{1+tan^2(a)}~da

    = \int csc(a)~sec^2(a)~da

    = \int csc(a)~(1+tan^2(a))~da

    = \int csc(a)~da + \int \frac{sin(a)}{cos^2(a)}~da

    = -ln| csc(a)+cot(a) | + sec(a)~da

    = -ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\rig  ht|+(2e^{2t}+1)^{-1/2} + C

    I've done it twice and gotten this answer, but it just seems kind of ugly and unlikely, and my calculator told me a different answer
    Note that:

    1. \sec a = \sqrt{2 e^{2t} + 1}, NOT (2e^{2t}+1)^{-1/2} = \frac{1}{\sqrt{2 e^{2t} + 1}}:

    \tan a = \sqrt{2} e^t \Rightarrow \tan^2 a = 2 e^{2t} \Rightarrow 1 + \tan^2 a = 1 + 2 e^{2t} \Rightarrow \sec^2 a = 2e^{2t} + 1.


    2. \ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\ri  ght| = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(\sqrt{2} e^t)

     = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(e^t) - \ln \sqrt{2} = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t - \ln \sqrt{2}.


    The \ln \sqrt{2} can be soaked up into C. Then you get an answer that will almost agree with what your calculator is probably telling you ...... The snag is that there shouldn't be a negative in front of you log .....

    In other words, the answer is


    \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t + \sqrt{2 e^{2t} + 1} + C


    NOT -\ln \left( 1+\sqrt{2e^{2t}+1} \right) + t + \sqrt{2 e^{2t} + 1} + C.


    I can't see off-hand where your unwanted negative has come from ......
    Last edited by mr fantastic; March 3rd 2008 at 02:55 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    723
    Awards
    1
    Quote Originally Posted by mr fantastic View Post
    Note that:

    1. \sec a = \sqrt{2 e^{2t} + 1}, NOT (2e^{2t}+1)^{-1/2} = \frac{1}{\sqrt{2 e^{2t} + 1}}:

    \tan a = \sqrt{2} e^t \Rightarrow \tan^2 a = 2 e^{2t} \Rightarrow 1 + \tan^2 a = 1 + 2 e^{2t} \Rightarrow \sec^2 a = 2e^{2t} + 1.


    2. \ln\left|\frac{1+\sqrt{2e^{2t}+1}}{\sqrt{2}e^t}\ri  ght| = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(\sqrt{2} e^t)

     = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - \ln(e^t) - \ln \sqrt{2} = \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t - \ln \sqrt{2}.


    The \sqrt{2} can be soaked up into C. Then you get an answer that will almost agree with what your calculator is probably telling you ...... The snag is that there shouldn't be a negative in front of you log .....

    In other words, the answer is


    \ln \left( 1+\sqrt{2e^{2t}+1} \right) - t + \sqrt{2 e^{2t} + 1} + C


    NOT -\ln \left( 1+\sqrt{2e^{2t}+1} \right) + t + \sqrt{2 e^{2t} + 1} + C.


    I can't see off-hand where your unwanted negative has come from ......
    Thank you, it's been a very long couple of weeks >.<

    the secant thing, I just made a careless error on, upon review, I see your answer is correct.

    And good eye on pulling out the t in the denominator, that's definitely why my answer was so far off from theirs. You'd think I'd remember to check such things :/


    The negative on the derivative of csc(a) came because I don't have that function's integral memorized, so I just went to look it up by typing it into the calculator. It told me:

    \int csc(x) = -\ln | cot(x) + csc(x) | + C

    out of annoyance with myself for not just figuring it out, I went to figure it out just now... and I'm having a tough time of it, can you show me how to it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by angel.white View Post
    Thank you, it's been a very long couple of weeks >.<

    the secant thing, I just made a careless error on, upon review, I see your answer is correct.

    And good eye on pulling out the t in the denominator, that's definitely why my answer was so far off from theirs. You'd think I'd remember to check such things :/


    The negative on the derivative of csc(a) came because I don't have that function's integral memorized, so I just went to look it up by typing it into the calculator. It told me:

    \int csc(x) = -\ln | cot(x) + csc(x) | + C

    out of annoyance with myself for not just figuring it out, I went to figure it out just now... and I'm having a tough time of it, can you show me how to it?
    Go to the relevant link here for the easy way. The tougher way uses the Weiestrass substitution.

    I know why you have the negative, the thing is that it's not wanted in the correct answer ...... Can't see off-hand why it's there when it shouldn't be ......
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Why make this thing so hard?

    Quote Originally Posted by angel.white View Post
    Can anyone do this integral for me? I am not convinced my answer is correct :/

    \int \sqrt{1+2e^{2t}}~dt
    Substitute z^2=1+2e^{2t}, the integral becomes \int {\frac{{z^2 }}<br />
{{z^2  - 1}}\,dz}  = z + \frac{1}<br />
{2}\ln \left| {\frac{{z - 1}}<br />
{{z + 1}}} \right| + k.

    Just back substitute.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Krizalid View Post
    Why make this thing so hard?


    Substitute z^2=1+2e^{2t}, the integral becomes \int {\frac{{z^2 }}<br />
{{z^2  - 1}}\,dz}  = z + \frac{1}<br />
{2}\ln \left| {\frac{{z - 1}}<br />
{{z + 1}}} \right| + k.

    Just back substitute.
    Yes, I'd thought of that. And of course an answer in this form immediately suggests a simplification using the definition of the inverse tanh function ....

    But hard or easy, all valid methods should lead to equivalent answers.

    I generally like to find the mistake in the method used by the person asking the question (unless the method follows the Highway to Hell) before suggesting other approaches. I find that gives the person greater ownership of the solution .....
    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