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Sub in the two intervals?

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int[/tex][tex]^{1/\sqrt{2}}_{0}[/tex] [tex]\stackrel{arcsinx}{\sqrt{1-x^2}}[/tex]


    2. Relevant equations

    The 0 is supposed to be on the bottom of the intergal, but I could not format it to go there.

    3. The attempt at a solution
    My attempt was to set u= arcsinx, then DU would equal the bottom, so it would be [tex]\stackrel{u}{du}[/tex]...but from there I get stuck. I know the answer is supposed to be .308...Help anyone?
     
    Last edited by a moderator: Jun 1, 2014
  2. jcsd
  3. Sep 21, 2009 #2
    Re: Integration

    [tex]\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx[/tex]

    \int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}

    If you set u=arcsinx

    [tex]du = \frac{dx}{\sqrt{1-x^2}}[/tex]

    [tex]\int u du = \int {arcsinx\over \sqrt{1-x^2}} dx[/tex]
     
  4. Sep 21, 2009 #3
    Re: Integration

    Ok, that makes sense. Then from there do you just sub in the two intervals?
     
  5. Sep 21, 2009 #4
    Re: Integration

    Yeah you need to make sure the intervals are consistent i.e. u=arcsinx, sinu=0 or 1/sqrt(2) quite simple, and that way you wont need to change all back to x.
     
  6. Sep 21, 2009 #5
    Re: Integration

    If possible, could you set up the substitution for me so I can see how it looks, and then I can work from that?
     
  7. Sep 21, 2009 #6
    Re: Integration

    [tex]\int_0^{1\over\sqrt{2}}\frac{\arcsin(x)}{\sqrt{1-x^2}}dx[/tex]

    using the substition:

    [tex] u = \arcsin(x) [/tex]

    [tex] {du\over dx} = {1\over \sqrt{1-x^2}} \Rightarrow du = {dx\over \sqrt{1-x^2}} [/tex]

    [tex] \int u du = \int \arcsinx {dx\over \sqrt{1-x^2}} [/tex]


    then make sure the limits are changed:


    [tex]x_0 = 0[/tex], [tex] x_1 = {1\over\sqrt{2}} [/tex]

    [tex]\arcsin(0) = u_0 = 0[/tex]

    [tex]\arcsin({1\over\sqrt{2}}) = u_1 = {\pi\over 4}[/tex]

    [tex]\int_0^{1\over \sqrt{2}} {\arcsin(x)\over \sqrt{1-x^2}} dx = \int_0^{\pi\over 4} u du [/tex]
     
  8. Sep 21, 2009 #7
    Re: Integration

    ok so


    [tex]\int^{\Pi/4}_{0}[/tex] [tex]dx/\sqrt{1-x^2}[/tex]


    Then from there I substitute pie/4 in for x?
     
  9. Sep 21, 2009 #8
    Re: Integration

    or would it be something like this..arcsin (Pie/4) - arcsin (0)??
     
  10. Sep 21, 2009 #9
    Re: Integration

    No. The whole integral has been transformed.

    [tex] \int_{0}^{\pi\over 4} u du = \int_{0}^{1\over\sqrt{2}} \frac{\arcsin (x)}{\sqrt{1-x^2}} dx [/tex]

    Basically because you swapped x for sin(u) (essentially) if x is [tex]1\over\sqrt{2}[/tex] then u has to be [tex]\pi\over 4[/tex] doesn't it? So everything is consistent? 0 stays the same because sin(0)=0.
     
  11. Sep 21, 2009 #10
    Re: Integration

    True, so we have that set up. Now where do we go from there?
     
  12. Sep 21, 2009 #11
    Re: Integration

    integrate u from 0 to pi/4 and thats it.
     
  13. Sep 21, 2009 #12
    Re: Integration

    so I would get .903??

    The answer in the back of the book for this problem says .308.
    Did I do something wrong?
     
  14. Sep 22, 2009 #13
    Re: Integration

    on calcchat.com, they have something like this for their new integral.

    [tex]\int^{\frac{1}{\sqrt{2}}}_{0}[/tex] [tex]\frac{1}{2}[/tex] arcsin2x


    then from there they get [tex]\frac{\Pi^2}{32}[/tex]


    which approximates to 0.308, but how on earth did they get that new integral? Does what you see here make any sense?
     
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