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Homework Help: Trig anti-dervative

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Int of sin x over - sin ^2 x of dx

    3. The attempt at a solution

    I don;t know if I have the right question. But I just couldn't reproduce the problem (from my answer)!!
    please take a look at my note, see if you can reproduce the original question

    i am talking about question #7

    Attached Files:

  2. jcsd
  3. Mar 11, 2009 #2


    Staff: Mentor

    Your attachment is still pending approval.
    Is this your integral?
    [tex]\int \frac{sin x}{-sin^2 x}dx[/tex]

    If so, this is the same as
    [tex]\int -csc (x) dx[/tex]
    If I recall correctly, this can be done using integration by parts.
  4. Mar 11, 2009 #3
    yeah i think that;s what it is
    but if we use -csc x can it be done easily?
    i mean cscx does not have any Anti-derv...
  5. Mar 11, 2009 #4


    Staff: Mentor

    But -sin^2(x) isn't equal to cos^2(x), so your first step is incorrect.
  6. Mar 11, 2009 #5
    then shouldn't it be int of sinx times -sin^2 x?
  7. Mar 11, 2009 #6


    Staff: Mentor

    I don't know what you mean. This is what you wrote in your first post in this thread:
    By "over" I assume you mean the quotient of sin(x) and -sin^2(x), which is what I showed in the integral.
  8. Mar 11, 2009 #7
    i am sorry, i am referring to your #4
    you said my first step was wrong

    now, i knew the mistake, and this is what i did

    original question:
    Int of sin x over - sin ^2 x of dx

    first step, change the bottom, -sin^2 (x) to this form --> 1 over csc^2 (x)

    so the entire int will become

    int of sin (x) times csc^2 (x)

    because 1/sinx = cscx, then 1 / -sin^2 (x) = csc^2 (x)
    am i correct?
  9. Mar 11, 2009 #8


    Staff: Mentor

    Or, using inline LaTeX tags,
    [itex]\int sin(x)/(-sin^2(x)) dx[/itex]
    You've lost a sign. -sin^2(x) = -1/csc^2(x)
    Or [itex]\int sin(x)(-csc^2(x)) dx[/itex]
    1 / -sin^2 (x) = -csc^2 (x)
  10. Mar 11, 2009 #9
    You're going around in circles. The integrand simplifies to -1/sin x or -csc x.

    It is not at all obvious how to integrate csc x. Try multiplying the integrand by a certain factor which allows you to make a substitution, but which doesn't change the value of the integrand. Hint: the final result involves a logarithm.
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