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Two different results depending on integration technique

  1. Jul 4, 2011 #1
    Hi,

    I'm having peculiar difficulties with a rather easy integral, namely the integral of -sin(x) / (cos(x))^3. The problem is that depending on which integration technique I choose, I end up with two different result. Moreover, Mathematica gives me one of these two results, while the solution book gives me the other. But the two cannot possibly be the same, unless I'm really overlooking something.

    Method One: substitution

    u = cos(x)
    du = -sin(x) dx

    [itex]\int[/itex]-sin(x)/cos^3(x) = [itex]\int[/itex]u^(-3) = -1/2 u^(-2) = -1/2 sec^2 x

    That is also the result that Mathematica gives me

    Method Two: Trig Identities, then substitution

    -sin x / cos^3 x = -tan x * sec^2 x

    u = tan x
    du = sec^2 x dx

    [itex]\int[/itex]-sin(x)/cos^3(x) = - [itex]\int[/itex]u = -1/2 u^2 = -1/2 tan^2 x

    That's the result that the solution book gives.

    It is obvious that the two resulting functions are different. What is wrong here?

    Thanks,
    Alex
     
  2. jcsd
  3. Jul 4, 2011 #2
    Sorry to say that both results are false because the constants have been forgotten.
    If the constants are there, both results are correct and identical :
    -(1/2)sec^2 (x) +C1 = -(1/2)tan^2 (x) +C2
    C1 = (1/2) + C2
     
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