calculus

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Find the indefinite integral of 1-tanx/1+tanx

Dont know how to really approach this question. Should i use identities, or is there a power series i can use?

  • calculus -

    There are standard formulae for integrals of rational functions of trigonometric formulae. In this case, you can simplify things as follows.

    Let's use the abbreviation:

    t = tan(x)

    s = sin(x)

    c = cos(x)

    We can write:

    (1-t)/(1+t) =

    (c-s)/(c+s) =

    (c - s)^2/(c^2 - s^2) =

    (c^2 + s^2 - 2cs)/(c^2 - s^2)

    Then use the trigonometric identities:

    c^2 + s^2 = 1

    2 cs = sin(2x)

    c^2 - s^2 = cos(2x)

    to obtain:

    (1-t)/(1+t) =

    1/(cos(2x)) - tan(2x)

    Integrating tan(2x) is trivial. You can integrate 1/cos(2x) e.g. by putting
    x = (pi/4 - u), so that cos(2x) =
    sin(2u. Then

    1/sin(2u) =

    [cos^2(u) + sin^2(u)]/[2 sin(u)cos(u)] =

    1/2 [cot(u) + tan(u)]

    which is trivial to integrate.

  • calculus -

    how did you get from (c-s)/(c+s) to

    (c - s)^2/(c^2 - s^2)?

  • calculus -

    Multiply numerator and denominator by (c-s).

  • calculus -

    thanks

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