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Trig And calculus

  1. Oct 25, 2004 #1
    Need help in integration and trig
    How do you integrate
    1. 1/(1-x^5)
    2. 1/(1+x^4)

    and the trig question.
    Show that

    (a^2 - b^2)/c^2 = sin(A-B)/sin(A+B)
  2. jcsd
  3. Oct 25, 2004 #2


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    Every polynomial, such as 1- x5 and 1+ x4 can be factored, using real numbers, into a product of linear or quadratic terms.
    To factor 1- x5, find all complex roots to x5= 1. One is, of course 1 the others are complex conjugates which can be paired to give two quadratic factors. Then use "partial fractions".
    Same for 1/(1+x4).

    In the trig question, are we to assum that a, b, and c are lengths of sides opposite angles A, B, C? In a right triangle or general triangle?
  4. Oct 26, 2004 #3
    general triangle!
    a,b,c are length's
    A,B,C are opposite angles

    Sorry but i made a mistake in the first integration question
    its suppose to be 1/sqrt(1-x^5)

    When i used this integration in mathematica 5 : i got something like hypergeometric2f1....

    Can u help me out!
  5. Oct 26, 2004 #4


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    The 5 "fifth roots of unity" lie on a circle, in the complex plane of radius 1, equally spaced around the circle. The angle between them is 360/5= 72 degrees so they are;
    1, cos(72)+ i sin(72), cos(144)+ i sin(144), cos(216)+ i sin(216), cos(288)+ i sin(288).

    Since cos(72)= cos(288), sin(72)= -sin(288), cos(144)= cos(216), and sin(144)= sin(216), these are in pairs of complex conjugates (as they have to be in order to satisfy and equation with real coefficients.

    The solutions to x5= 1 are: 1, cos(72)+ i sin(72), cos(72)- i sin(72), cos(144)+ i sin(144), cos(144)- i sin(144) and so
    1- x= -(x-1)(x- cos(72)+ i sin(72))(x- 72- i sin(72))(x- cos(144)+ isin(144))(x- cos(144)- i sin(144))= -(x-1)((x-cos(72))2+ sin2(72))((x-cos(144)2+ sin2(144))

    = -(x-1)(x2- 2cos(72)+ 1)(x2-2cos(144)+ 1).

    Once you have that factorization you can expand 1/(1- x5) in partial fractions.
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