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Homework Help: Integration math course work

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the general indefinite integral of sinx/1-(sinx)^2



    3. The attempt at a solution

    I arrived at tanxsecx(dx), rewrote it as sinx(1/cos^2 x) = cosxtanx = sinx

    However, I know that the correct answer is secx. WHY? Can anyone explain why using trig functions (ie, without using u as a substitution)?
     
  2. jcsd
  3. Oct 25, 2007 #2

    Dick

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    sin(x)/cos(x)^2 isn't equal to cos(x)*tan(x).
     
  4. Oct 25, 2007 #3

    dynamicsolo

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    Don't simplify it: just leave it as (sin x) / (cos^2 x) . Now, could you use a u-substitution on

    [tex]\int \frac{sin x}{cos^{2} x} dx [/tex]?
     
  5. Oct 25, 2007 #4
    Why can't it equal if you substitute?
    sinx/(cosx)(cosx)
    (sinx)(1/(cosx)(cosx)
    the antiderivative of that gives...
    (cosx)(secx)(secx)
     
  6. Oct 25, 2007 #5
    Is there any way to get to the answer without substituting? What is making my method invalid?
     
  7. Oct 25, 2007 #6

    HallsofIvy

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    Because "tan(x) cos(x)" is equal to (sin(x)/cos(x))(cos(x))= sin(x). You have sin(x)/cos^2(x). The "second" cosine is in the denominator, not the numerator. That is (sin(x)/cos(x))(1/cos(x)= tan(x)sec(x). The integral of that is sec(x)+ C= 1/cos(x) + C which is exactly what you get if you leave it as sin(x)/cos^2(x) and make the substitution u= cos(x).
     
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