Results 1 to 4 of 4

Math Help - apostol calculus, trouble simplifying integral

  1. #1
    Member
    Joined
    Aug 2008
    Posts
    249

    apostol calculus, trouble simplifying integral

    i'm reading apostol's calculus book vol.1 and for those of you who have it i am having trouble on page 265, where he does the integral of 1 / (sinx + cosx) using a wierstrauss substitution. i understand the substitutions he made and the answer he got which was (sqrt(2) / 2) log[(tan(x/2) - 1 + sqrt(2)) / (tan(x/2) - 1 - sqrt(2))] + C. next he tries to simplify it by saying sqrt(2) - 1 = tan(pi/8). so the numerator can be simplified to tan(x/2) + tan(pi/8). i know he is going to use tan(a+b) = (tana + tanb) / (1 - tana tanb) to simplify so he wants the denominator to be in the form 1 - (tan a)(tan b).

    however, i am having trouble following his steps to get the denominator to that form. he does this in the book: | tan(x/2) - 1 - sqrt(2) | = (sqrt2 + 1)(sqrt2 - 1)tan(x/2) - 1 = (sqrt2 + 1) (1 - tan(x/2)tan(pi/8).

    i have no idea how he got from the original denominator which was tan(x/2) - 1 - sqrt(2) to (sqrt2 + 1)(sqrt2 - 1)tan(x/2) - 1. i expanded out the 2nd term in the equality but it did not match the 1st term in the equality. i am also confused on notation. on the first equality is that an absolute value sign? going from the first term to the second term | tan(x/2) - 1 - sqrt(2) | = (sqrt2 + 1)(sqrt2 - 1)tan(x/2) - 1 has really got me confused. please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    tan(x/2)=(sqrt2 + 1)(sqrt2 - 1)tan(x/2)

    NOW:

    tan(x/2) - 1 - sqrt(2) = (sqrt2 + 1)(sqrt2 - 1)tan(x/2) + (- 1 - sqrt(2))= (sqrt2 + 1){(sqrt2 - 1)tan(x/2) -1}

    Now:
    tan(pi/8)=sqrt2 - 1
    (It's one solution of x^2+2x-1=0 if you interested)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    tan(pi/8)=sqrt2 - 1
    ====================

    Recall that tan (pi/4) = 1.

    put pi/8 into your first equation.

    Then

    tan( 2*pi/8 ) = 2 tan (pi/8) / [ 1 - tan^2 (pi/8) ] ==>

    tan( pi/4 ) = 2 tan (pi/8) / [ 1 - tan^2 (pi/8) ] ==>

    1 = 2 tan (pi/8) / [ 1 - tan^2 (pi/8) ] ==>

    Let x = tan (pi/8) so I can see the forest from the trees....

    So

    1 = 2x / [ 1 - x^2 ] ==> 1 - x^2 = 2x ==> T^x + 2x = 1

    Solve by completing the square:

    (x + 1)^2 = 1 + 1 ==> (x + 1)^2 = 2 ==> x = sqrt(2) -1.

    And x = tan (pi/8). So tan (pi/8) = sqrt(2) -1.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2008
    Posts
    249
    excellent explanation! and thank you for also explaining how tan(pi/8) = sqrt(2) - 1.

    very helpful thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: March 10th 2013, 06:13 PM
  2. Replies: 3
    Last Post: October 7th 2010, 06:28 AM
  3. [SOLVED] Having trouble simplifying this integral
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 3rd 2010, 01:36 PM
  4. [SOLVED] Simplifying Expressions, having trouble with this one
    Posted in the Algebra Forum
    Replies: 8
    Last Post: September 1st 2010, 11:00 AM
  5. Trouble Simplifying/Factorising
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 25th 2009, 05:09 AM

Search Tags


/mathhelpforum @mathhelpforum