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Math Help - Basic integration problem, general power formula

  1. #1
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    Basic integration problem, general power formula



    Hi. I'm wondering how come the answer has an 8 in the denominator. My answer has 2, although the nominator remains the same?

    u = (sin(4x)^-1)
    du = (1 - (16x^2))^(-1/2)

    n = 1
    n + 1 = 2

    any hints, thanks?
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  2. #2
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    Re: Basic integration problem, general power formula

    Quote Originally Posted by togo View Post


    Hi. I'm wondering how come the answer has an 8 in the denominator. My answer has 2, although the nominator remains the same?

    u = (sin(4x)^-1)
    du = (1 - (16x^2))^(-1/2)

    n = 1
    n + 1 = 2

    any hints, thanks?
    I'm not completely sure you have the method. Just to verify, you are using the substitution x = \frac{sin( \theta )}{4} right? In that case the integral turns into
    \int \frac{sin^{-1} ( sin( \theta ))}{4}d \theta = \int \frac{ \theta }{4} d \theta

    where the extra 2 comes from the integration.

    -Dan
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  3. #3
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    Re: Basic integration problem, general power formula

    I would prefer to make substitution sin^-1 ( 4x) = t that gives [1/sqrt(1-16 x^2) ] *4 dx =dt Thus we have [1/sqrt(1-16 x^2) ] dx =dt/4
    now the integral becomes t/4 dt. Now you can take it further.
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