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3 Integration questions

  1. Oct 4, 2004 #1
    (1) [tex]\alpha(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{a^ 2-u^2}} [/tex]

    (2) [tex]\beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^ 2+b}} [/tex]

    (3) [tex]\beta(t-t_{0})=\int_{R_{0}}^{R(\Theta)}\frac{du}{u\sqrt{u^ 2-b}} [/tex]

    Should I use trig subs? If so, what should my "u" be? :frown:
  2. jcsd
  3. Oct 6, 2004 #2
    i've not done integrals with the limits and equalities such as you have posted, but, the definite integrals are easy enough.

    use the trig identities sin^2 x + cos^2 x = 1

    for the first problem, let a = u sin x, da = u cos x
    from there, you can see that you can factor out a u^2 in the radical sign. you are left with sin^2 x - 1, which equals cos^2 x. the square root of cos^2 x is cos x.
    now you should have definite integral of 1/(sin^2 x cos x)
    use power reduction to simpliy sin^2 x in terms of cos. i gotta go to class now, sorry. if nobody has gotten to it in 4 hours from now, i'll be back. and also learn the tex commands so this is readable
  4. Oct 6, 2004 #3
    for the first one use [tex]u=asin(x)[/tex] or [tex]u=acos(x)[/tex]
    second one [tex]u=b^{1/2}tan(x)[/tex]
    third [tex]u=b^{1/2}sec(x)[/tex]

    you must be very carefull with the sign of the functions thoug, remember that
  5. Oct 6, 2004 #4
    thank you!
  6. Oct 7, 2004 #5


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    In general, if you see [itex]\srqt{1- x^2}[/itex] you should immediately think "cos2= 1- sin2".

    If you see [itex]\sqrt{1+ x^2}[/itex] you should immediately think "1+ tan2= sec2".

    If you see [itex]\sqrt{x^2- 1}[/itex] you should immediately think "sec2- 1= tan2".
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