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Homework Help: U substitution

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve using u-substitution.

    [tex] \int \frac{1}{sin x} dx [/tex]

    3. The attempt at a solution
    I looked up the integral answer and I can see the connection of the csc, but I don't know how to get this to be du/u for the ln function.
     
  2. jcsd
  3. Dec 8, 2009 #2
    That's the last thing you should do when solving problems as it defeats the purpose of learning via solving problems.

    There is no need to get things right after just one substitution. You can do substitution after substituton as many times you like. So why not start with putting sin(x) = u and see what you get first?
     
  4. Dec 8, 2009 #3
    I already did all that, it's just too much to list. I tried u = sin x at first but didn't get very far. Then I thought that since the sin is on the bottom I could try cos x since -sin is derivative of cos but didn't get very far again. Then I tried rewriting 1/sin x using trig identities but I couldn't find any that would help. (:

    EDIT: I have a question though. When I was doing u = sin x, du = cos x dx. So would this be correct?

    [tex] \int \frac{1}{u}cos x dx [/tex]

    [tex] csc x \int \frac{du}{u} [/tex]

    [tex] csc x \ln sin x [/tex]
     
    Last edited: Dec 8, 2009
  5. Dec 8, 2009 #4
    Let's start again then. If you put

    sin(x) = u

    you get:

    dx/sin(x) = du/u 1/sqrt(1-u^2)

    (up to a plus/minus sign which can be ignored for the moment). Do you agree?

    Edit, I see your latest edit now. You have to express cos(x) in terms of u.
     
  6. Dec 8, 2009 #5
    I'm not following what you have written there, particularly I don't see where you're getting 1/sqrt(1-u^2) from. Isn't that a derivative of the arcsin?
     
  7. Dec 8, 2009 #6
    If you put

    sin(x) = u

    then

    x = arcsin(u)

    and thus:

    dx = dx/du du = 1/sqrt(1-u^2) du

    You can also say that:

    sin(x) = u ------->

    d[sin(x)] = du -------->

    cos(x) dx = du ---------->

    dx = 1/cos(x) du = 1/sqrt[1-sin^2(x)] du = 1/sqrt(1-u^2) du

    So, we have:

    dx/sin(x) = 1/u 1/sqrt(1-u^2) du
     
  8. Dec 8, 2009 #7
    Ok, what's the next step?

    edit: would you integrate after that? it looks like you'd have the natural log of a mess :P
     
  9. Dec 8, 2009 #8
    Put u = 1/v. The idea is that the du/u part remains the same (up to a sign)

    du = -1/v^2 dv

    1/u = v

    du/u = -1/v dv

    but

    1/sqrt(1-u2)

    becomes

    1/sqrt(1-1/v^2)

    so, you can bring the v in the denominator inside the square root and end up with something simple.
     
  10. Dec 8, 2009 #9
    gotcha, thanks
     
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