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Homework Help: Definite Integral

  1. Aug 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Find using substitution [tex]x=\sin \theta[/tex]

    [tex]\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]\int _0^{0.5}\frac{x}{\sqrt{1-x^2}}dx[/tex]

    [tex]=\int_0^{\frac{\pi}{6}}\sin \theta\;d\theta[/tex]


    What I want to ask is about changing the upper bound of the integral.

    For x = 0.5 :
    [tex]0.5 = \sin \theta[/tex]

    Here I choose [tex]\theta = \frac{\pi}{6}[/tex]. But what if I choose [tex]\theta = \frac{5}{6}\pi\;??[/tex]

    So, the integral :
    [tex]=\int_0^{\frac{5}{6}\pi}\sin \theta\;d\theta[/tex]


  2. jcsd
  3. Aug 29, 2009 #2
    A good way to check your answers on integration problems is to solve them in different ways and compare their answers. One way to do this is to use another integration method.
  4. Aug 29, 2009 #3


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    Be careful!

    We have, ignoring boundary values:

    Do you see what you need to be careful about when simplifying this further?

    In particular, why is your choice: "But what if I choose [itex]\theta=\frac{5}{6}\pi[/itex]" troublesome?
  5. Aug 29, 2009 #4
    Hi arildno and espen180
    In particular, why is your choice: "But what if I choose [itex]\theta=\frac{5}{6}\pi[/itex]" troublesome?[/QUOTE]

    Ah, maybe you are referring to the square root. The value of [tex]\cos \theta[/tex] should be positive so the angle should be in first quadrant, where the values of sin and cos are positive.

    I've checked it using substitution u = 1-x^2 and got the answer. What I am confused is about why I can't take the upper bound to be (5 pi)/6. But I think I get it now

    Thank you very much, arildno and espen180 :smile:
  6. Aug 29, 2009 #5
    It is same either if you pick п/6 or 5п/6 because п - п/6 = 5п/6 and both result with positive value of sin(x).

    What arildno meant to say is that you should be careful with the absolute value [tex]\int{sin(x)\frac{cos(x)}{|cos(x)|}dx}=\int{sin(x)\frac{cos(x)}{\pm cos(x)}dx}=\int{\pm sin(x)dx}[/tex]

    So, you should consider the negative values for cos(x) and not just the positive ones. :smile:

    P.S And why you chose x=0 for sin(x)=0, why you did not choose x=п, since both are valid?
  7. Sep 6, 2009 #6
    Hi njama

    I don't really understand about the absolute sign. Why can it be negative? I think it's should be positive since it's a square root and we don't have to consider the negative value.

    Thanks :smile:
  8. Sep 6, 2009 #7
    Ok, I will explain.

    For example if you have [itex]y^2=16[/itex] so that we can write |y|=4

    What that means is y can be either -4 or 4 so that |y|=4 or [itex]y=\pm 4[/itex]

    Lets check it out:

    (-4)2=16, 42=16 or |-4|=4 but also |4|=4

    In this case take cos(x) = y, and you will see what I am talking about. :smile:
  9. Sep 6, 2009 #8


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    Of course, [itex]|cos(\theta)|[/itex] cannot be negative. But you can write it as [itex]\pm cos(\theta)|[/itex] because [itex]cos(\theta)[/itex] itself can be positive or negative.

    IF [itex]\theta= (5/6)\pi[/itex] then [itex]cos(\theta)= cos((5/6)\pi)= -\sqrt{3}/2[/itex] so [itex]|cos(\theta)|= - cos(\theta)= \sqrt{3}/2[/itex].
  10. Sep 6, 2009 #9
    Hi njama and Mr. HallsofIvy

    This is my interpretation after reading your posts.

    [tex]\int_{0}^{0.5}\frac{xdx}{\sqrt{1-x^{2}}}[/tex] can turn out to be:

    [tex]1.\int_{0}^{\frac{\pi}{6}}\sin \theta\;d\theta\;\text{or}[/tex]

    [tex]2.\int_{\pi}^{\frac{5}{6}\pi}-\sin \theta\;d\theta[/tex]

    Am I right? Or maybe there are any other forms with different upper and lower bounds?

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