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Math Help - inverse trig integration:two methods=two diff ans HELP!!

  1. #16
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    Quote Originally Posted by i_zz_y_ill View Post
    when u take 4 out side with the 3/4,,,,,,,,(this is jhevs method), doesnt it become
    3/4 (integral) 1/((3x+1)^2)/4 + 1

    not 3/3 (integral) 1/((3x+1)^2)/2 +1
    It should read 1/ [(3x+1)/2]^2 + 1
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  2. #17
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    when u take 4 out side with the 3/4,,,,,,,,(this is jhevs method), doesnt it become
    3/4 (integral) 1/((3x+1)^2)/4 + 1

    not 3/3 (integral) 1/((3x+1)^2)/2 +1
    ok, look carefully at what i did. i never typed \frac {(3x + 1)^2}2, i typed \left( \frac {3x + 1}2  \right)^2. BIG difference

    \frac {(3x + 1)^2}4 = \frac {(3x + 1)^2}{2^2} = \left( \frac {3x + 1}2 \right)^2
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  3. #18
    Member i_zz_y_ill's Avatar
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    uhuh

    k look at this example, i shud be able to do this without substitution....
    using (INT) 1/(a^2 + (f(x))^2 ) = 1/a arctan x/a

    i can say that (int) 4/(x^2 -2x + 3)dx = 4(int) 1/(x-1)^2 +2 dx
    there = 4/root2 arctan (x-1)/root2 + c

    which is correct,, are you saying thatbecause with my previous example the
    f(x) has a coefficient if you like of three before it i have to use substitution?
    Last edited by mr fantastic; May 30th 2009 at 05:13 AM. Reason: Removed bad language
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  4. #19
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    See my signature for LaTeX typesetting, really, see it to make your posts clearer.
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  5. #20
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    Quote Originally Posted by i_zz_y_ill View Post
    k look at this example, i shud be able to do this without substitution....
    using (INT) 1/(a^2 + (f(x))^2 ) = 1/a arctan x/a

    i can say that (int) 4/(x^2 -2x + 3)dx = 4(int) 1/(x-1)^2 +2 dx
    there = 4/root2 arctan (x-1)/root2 + c

    which is correct,, are you saying thatbecause with my previous example the
    f(x) has a coefficient if you like of three before it i have to use substitution if so then BOLLOCKS to that.. what i actually mean is BOLLOCKS TO THAT!!!!!
    ????
    Nope, you don't have to use substitution if you can remember to deal with the coefficient later on (i.e. divide by 3). Jhevon did a great job illustrating the basic principles behind his solution, and it's very comprehensive - but ultimately you don't have to use substitution if you know how to solve it otherwise.
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  6. #21
    Member i_zz_y_ill's Avatar
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    I Totally Gottit Triumph

    U HAVE TO SUBSTITUTE U=3X+1
    DX=DU/3 CANCELS + GIVES ANSWER PRESUMABLY I MUST SUBSTITUTE WHENEVER THERE A COEFFICIENT OF X IN THOSE BRACKETS!?
    Last edited by mr fantastic; May 30th 2009 at 05:14 AM.
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  7. #22
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    Effectively, you could put it that way, yes, or you could just do it in your head =)
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