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Math Help - Find the anti-derivative

  1. #1
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    Find the anti-derivative

    integration of t^2/9+t^6 dt
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    Quote Originally Posted by simsima_1 View Post
    integration of t^2/9+t^6 dt
    this is a substitution problem (you may want to do a substitution twice, but you don't have to). let u = t^3
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    if i use t^3 as my u will that affect the (9+t^6)?
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    Quote Originally Posted by simsima_1 View Post
    if i use t^3 as my u will that affect the (9+t^6)?
    yes. 9 + t^6 = 9 + (t^3)^2 = 9 + u^2
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  5. #5
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    ok...now i have (du/3)/9+u^2 and i have no idea where to go next...i could pull out a (1/3) in the front but then what? i also know that inverse tan is 1+u^2 which is something i'm looking out for to make the equation work...thanks for all the help by the way...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by simsima_1 View Post
    ok...now i have (du/3)/9+u^2 and i have no idea where to go next...i could pull out a (1/3) in the front but then what? i also know that inverse tan is 1+u^2 which is something i'm looking out for to make the equation work...thanks for all the help by the way...
    you have \frac 13 \int \frac {du}{9 + u^2} so far

    now we want to get the denominator to the form 1 + x^2 so we can use the arctan for the integral. the 9 is the problem, so factor it out and simplify

    \frac 13 \int \frac {du}{9 + u^2} = \frac 13 \int \frac {du}{9(1 + u^2/9)}

    = \frac 1{27} \int \frac {du}{1 + (u/3)^2}

    now you may want to do a second substitution here of x = \frac u2, to get: \frac 2{27} \int \frac {dx}{1 + x^2}
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    is up to his old tricks again! Jhevon's Avatar
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    i made a typo in my last post. i typed let x = u/2 when it should be x = u/3. make the necessary corrections
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