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Calculus Problems part I

  1. Jan 11, 2006 #1
    Can someone show me the steps on finding:

    Note i dont have the answer for the last two questions. Im so confused help :surprised

    1. ∫1/cscx-1

    It says the answer is:

    _________ - x
    cot (x/2)-1

    2. ∫1/(1-sin x)
    It says the answer is:

    cot (x/2)-1

    3. ∫1/(1-e^-x)

    4. ∫cos x ln x
  2. jcsd
  3. Jan 12, 2006 #2


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    Homework Helper

    Yes, we can help you, but we won't show you the step-by-step solution!!! Read here!
    You should note that:
    ∫1/cscx-1 is not the same as ∫dx/(cscx-1).
    When seeing these kinds of integrals (i.e: trig integral), generally, what you should do is to use u-substitution (u = tan(x / 2)).
    Let [tex]u = \tan \left( \frac{x}{2} \right) \Rightarrow du = \frac{dx}{2 \cos ^ 2 \left( \frac{x}{2} \right)} = \frac{dx (1 + u ^ 2)}{2} \Rightarrow dx = \frac{2du}{1 + u ^ 2}[/tex]
    You'll also have:
    [tex]\sin x = \frac{2u}{1 + u ^ 2}[/tex]
    [tex]\int \frac{dx}{\csc x - 1} = \int \frac{\sin x \ dx}{1 - \sin x} = \int -\frac{-\sin x \ dx}{1 - \sin x} = \int -\frac{1 - \sin x - 1 \ dx}{1 - \sin x}[/tex]
    [tex]= -\int dx + \int \frac{dx}{1 - \sin x} = -\int dx + \int \frac{2du}{(1 + u ^ 2) \left( 1 - \frac{2u}{1 + u ^ 2} \right)} = -\int dx + \int \frac{2du}{(1 + u ^ 2) \left( 1 - \frac{2u}{1 + u ^ 2} \right)}[/tex].
    Can you go from here?
    It can be done exactly the same as the first one.
    To do some kinds of integrals like this, you can use the substitution u = ex.
    This can be done by integrating by parts. Looking closely at that, what should be u, and what should dv?
    You should note that:
    [tex]\int \frac{\sin x}{x} dx = \mbox{si} (x) + C[/tex].
    Can you go from here?
    If you get stuck somewhere, just should it out.
    Last edited: Jan 12, 2006
  4. Jan 12, 2006 #3


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    Staff Emeritus
    Science Advisor

    Strictly speaking [itex]\int1/csc x- 1[/itex] means [itex]\int\frac{1}{csc x}-1 dx[/itex] which is the same as [itex]\int sin(x)-1 dx[/itex] which is easy. I think you meant [itex]\int\frac{1}{csc x- 1}dx[/itex].
    That is the same as [itex]\int\frac{sin x}{1- sin x}dx[/itex]
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