Results 1 to 12 of 12

Math Help - Integration by substitution

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    73

    Integration by substitution

    I think that Integration by substitution is the appropriate method to use for the following function.

    \displaystyle\int\dfrac{2\sec x}{2+\tan x}

    I can simplify it to

    \displaystyle\int\dfrac{2}{2\cos x + \sin x}

    But I don't know if this is any better. Any ideas please

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You could try rewriting it as:

    \frac{2sec(x)}{2+tan(x)}=\frac{2}{\sqrt{5}cos(x-tan^{-1}(\frac{1}{2}))}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    73
    Hi

    Could you explain how you got the simplification please.

    James
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    You had \frac{2}{2cos(x)+sin(x)}. Which is cool.

    \frac{2}{\sqrt{5}(\frac{2cos(x)}{\sqrt{5}}+\frac{s  in(x)}{\sqrt{5}})}

    Now, note that cos(x-tan^{-1}(1/2))
    =cos(x)cos(tan^{-1}(1/2))+sin(x)sin(tan^{-1}(1/2))=\frac{2cos(x)}{\sqrt{5}}+\frac{sin(x)}{\sqrt{5  }}

    Just a matter of some algebra and noting an identity or two.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    73
    Thanks, I see where your coming from now.

    Sorry to ask another question but how should I handle the substitution?

    James
    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 bobred View Post
    I think that Integration by substitution is the appropriate method to use for the following function.

    \displaystyle\int\dfrac{2\sec x}{2+\tan x} \, {\color{red}dx}

    I can simplify it to

    \displaystyle\int\dfrac{2}{2\cos x + \sin x} \, {\color{red}dx}

    But I don't know if this is any better. Any ideas please

    Thanks
    Use the weierstrass substitution: The Weierstrass Substitution Example
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jan 2009
    Posts
    73
    Thanks Mr Fantastic for the reply, I've tried to follow the example several times to no avail, any tips would be appreciated.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    Just in case a picture helps...

    Spoiler:


    Straight continuous lines differentiate downwards, the straight dashed line likewise but with respect to the dashed balloon expression. So the triangular network on the right satifies the chain rule. The Weierstrass substitution is 't = tan of half the angle', from which are derived the formulae for sin x and cos x in terms of t.

    Don't integrate - balloontegrate!

    Balloon Calculus Forum
    Last edited by tom@ballooncalculus; January 25th 2010 at 01:43 PM. Reason: spoiler for pic which needed more room for missing step
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2009
    Posts
    73
    Hi

    Thanks for the responce, I feel so thick. How did you arrive at
    Attached Thumbnails Attached Thumbnails Integration by substitution-int.png  
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    Well, anti-clockwise from bottom-left, we do the Weierstrass subs for cos and sin (Mr Fantastic's link), simplify, then spot that

    \frac{2}{1 - t^2 + t}

    is like the bottom end of the standard result for tanh-1...



    So complete the square...
    Last edited by tom@ballooncalculus; June 19th 2009 at 02:50 AM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    ... and the resemblance...

    1 - t^2 + t

    = 1 - \large{[}(t - \frac{1}{2})^2 - \frac{1}{4}\large{]}

    = \frac{5}{4} - (t - \frac{1}{2})^2

    = \frac{5}{4} [1 - \frac{4}{5}(t - \frac{1}{2})^2]

    = \frac{5}{4} [1 - (\frac{2}{\sqrt5}t - \frac{1}{\sqrt5})^2]

    Don't forget you still have to revert from t to x.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jan 2009
    Posts
    73
    Hi

    Got it, thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2010, 11:12 PM
  2. Integration by Substitution Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 1st 2009, 08:28 PM
  3. integration by substitution xe^x^2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 5th 2008, 10:48 PM
  4. integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2008, 05:22 PM
  5. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 19th 2006, 03:46 PM

Search Tags


/mathhelpforum @mathhelpforum