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Math Help - antiderivative.

  1. #1
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    antiderivative.

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  2. #2
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    Quote Originally Posted by rcmango View Post
    I can't read that; int(sqrt(7 + ____), t = 0...1)dt

    Type what the integral is here please.
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  3. #3
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    I think I can read it correctly.

    In latex (when it works again) \int_{0}{1} \sqrt{9+4t^2}dx.

    In easier to read type: int sqrt(9+4t^2)dt, evaluated from 0 to 1.

    Before I do this, have you studied trig substitution, because if you haven't this might be slightly difficult to get at first?
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  4. #4
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    i know some trig identities if you mean that is trig substitution?
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  5. #5
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    Alright. Well draw a right triangle with the horizontal leg being 3 and the vertical leg being 2t. So it follows that the hypotenuse is sqrt(9+4^2).

    Now let's make some substitutions. Call the angle between the horizontal leg and the hypotenuse theta.

    tan(theta) = (2t)/3, and taking the derivative...

    sec^2(theta)d(theta) = (2/3)dt.

    So t = [3tan(theta)]/2 and dt = [3sec^2(theta)]/2

    Now plug this back into the original integral.

    int sqrt(9+4t^2)dt = int sqrt[(9/4)tan^2(theta)+9] *(3/2)sec^2(theta)d(theta).

    Is this making sense?
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