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Math Help - Integration, Substitution Rule

  1. #1
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    Integration, Substitution Rule

    I'm having trouble understanding what i'm doing here to get the answer.

    For instance

    Find indefinite integral for x^3cos((x^4)+2)dx

    ok first of all, what is the dx and why is it there at the end?

    To do this problem from the instructions, I let

    u = (x^4 ) + 2
    du = 4x^3 dx

    Is du just the derivative of u? Also why is dx there again and what does it represent?

    from here I would

    x^3 dx = (1/4)du

    Now that I have those, according to the substitution rule, it should be

    (cos u)((1/4)du) ok...now this is the part I don't understand even more is

    for Cos u, you find the antiderivative and get Sin u where u will be replaced with (x^4) + 2 giving you Sin((x^4) +2). That part is easily understandable. The part I don't understand is why is ((1/4)du) just = 1/4? where did the du go? U was replaced easily with ((x^4) +2) so why can't du be replaced with 4x^3 but instead it's replaced with 1/4? To my understanding, the Substitution Rule formula is f(u) du and this means f(u) times the derivative of u. so the equation should look like

    cos((x^4) + 2) multiplied by 4x^3. Apparently somewhere along in there i'm missing something and I don't get it. I'm interpreting the formula exactly the way it's written and can't get an understanding of what steps i'm skipping or missing.

    Any help is appreciated.
    Last edited by JonathanEyoon; April 18th 2008 at 01:47 PM.
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  2. #2
    Jen
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    So, we are trying to find the integral using a u du substitution. You chose the right u and du. so I will just walk through this...

    \int x^3 \cos (x^4+2) dx

    dx is called the differential it tells us what we are taking the integral with respect to x.

    So for the u du substitution we used u=x^4+2 \mbox{ and }du=4x^3dx

    So what we did here was we chose a u such that when we take it's derivative with respect to x (du/dx) we will be able to simplify our integral a bit using u and du.

    if we simplify the du to \frac{1}{4}du=x^3dx we can replace the x^3dx \mbox{ from the original integral with }\frac{1}{4}du

    substituting our u and du into the original integral we get

    \frac{1}{4} \int \cos u du I pulled the 1\4 out in front of the integral because we can do that with constants.

    which is a much simpler integral.

    Taking this intergal we get

    \frac{1}{4}\sin u +c \mbox{ now substitute our } x^4+2 \mbox{ back in for u }

    and we get

    \int x^3 \cos (x^4+2) dx = \frac{1}{4} \sin (x^4+2)+c
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  3. #3
    Jen
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    du is just the derivative of u and the reason that du doesn't get replaced back with 4x^3 is because it is gone after we take the integral.

    There are more technical ways to explain the differentials using reimman sums and what not, but I don't feel confident enough to attempt this.
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  4. #4
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    "the reason that du doesn't get replaced back with is because it is gone after we take the integral."


    I can't quite understand how it's gone after we take the integral.?



    if we simplify the du to we can replace the

    For the part above why do we have to do that?
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  5. #5
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    Thread in "Calculus" subform

    close thread please. Mistakenly double posted.
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