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Math Help - Integration technique help, please!

  1. #1
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    Integration technique help, please!

    What technique should I use to solve this one?

    int(16+x^2)^(-3/2)

    I tried a few techniques, including partial fractions, but I couldn't work it out. Can anybody help me?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by riverjib View Post
    What technique should I use to solve this one?

    int(16+x^2)^(-3/2)

    I tried a few techniques, including partial fractions, but I couldn't work it out. Can anybody help me?
    Start with the substitution:
    x = 4*tan(t)
    and note that
    1 + tan^2(t) = sec^2(t)

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by riverjib View Post
    What technique should I use to solve this one?

    int(16+x^2)^(-3/2)

    I tried a few techniques, including partial fractions, but I couldn't work it out. Can anybody help me?
    I agree with topsquark, trig substitution seems like the way to go.

    you have int(16+x^2)^(-3/2) = int((4)^2+x^2)^(-3/2) = int[1/(sqrt((4)^2 + x^2)^3].

    trig substitution says, if your integral contains sqrt(a^2 + x^2), where a is a constant, you use the sustitution, x = atan@ and the identity 1 + tan^2@ = sex^2@
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    1 + tan^2@ = sex^2@
    Uhh.....Heh heh heh heh. He said "sex" Beavis.

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by topsquark View Post
    Uhh.....Heh heh heh heh. He said "sex" Beavis.

    -Dan
    haha, yea, i did. it should be sec(x), haha, thats what trying to type fast gets you
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  6. #6
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    Thank you both! That makes perfect sense.
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  7. #7
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    Need a bit more help...sorry

    I started by using the substitution x=4tan@, but how do i handle the fact that a^2+x^2 is cubed?? none of the trig substitution problems we've done have added that kind of complication.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by riverjib View Post
    Int[(16+x^2)^(-3/2)dx]
    Let x = 4*tan(t) ==> dx = 4*sec^2(t) dt

    Then
    Int[(16+x^2)^(-3/2)dx] =
    Int[(16+16{tan^2(t)})^(-3/2) * 4sec^2(t) dt]

    = 4*(16)^(-3/2)*Int[(1 + tan^2(t))^(-3/2) sec^2(t) dt]

    = 4*(1/64)*Int[(sec^2(t))^(-3/2) sec^2(t) dt]

    = (4/64)*Int[sec^2(t)/sec^3(t) dt]

    = (1/16)*Int[1/sec(t) dt]

    = (1/16)*Int[cos(t) dt]

    = (1/16)*sin(t)

    Now, t = tan^{-1}(x/4), thus

    Int[(16+x^2)^(-3/2)dx] = (1/16)*sin[tan^{-1}(x/4)]
    (plus some constant of integration.)

    Now, imagine a right triangle with legs x and 4, then the hypotenuse will be sqrt(x^2 + 16). We have an angle t (which is tan^{-1}(x/4)) across from the side of length x. Then the sine of this angle will be: x/sqrt(x^2 + 16). So
    sin[tan^{-1}(x/4)] = x/sqrt(x^2 + 16)

    So finally:
    Int[(16+x^2)^(-3/2)dx] = (1/16)*x/sqrt(x^2 + 16)

    (Note: I finally found the error in my original post. This is now correct.)

    -Dan
    Last edited by topsquark; March 20th 2007 at 05:02 AM.
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  9. #9
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    Thank you so much!

    Thanks Dan! I read over your solution, gave it a few hours, and then worked it out on my own. Now I really understand every trig substitution problem in the book. I'm ready for my exam :-)

    I really appreciate all your help.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by riverjib View Post
    Thanks Dan! I read over your solution, gave it a few hours, and then worked it out on my own. Now I really understand every trig substitution problem in the book. I'm ready for my exam :-)

    I really appreciate all your help.
    You are very welcome! For the record I found the error in my post and have corrected it.

    -Dan
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