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  1. #1
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    Integral Help

    int(3*(x^2+9)^10)

    How can you do this using U sub?
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by anywho View Post
    int(3*(x^2+9)^10)

    How can you do this using U sub?
    \int 3(x^2+9)^{10}dx

    This is not a candidate for regular subsitution. There is no x in front of our bracket. In order to use regular substution whatever is in front of the bracket must be 1 order less then the term(s) in the bracket. In this case we would need an X.

    You cannt evaluate via this method.

    What you can do is integrate by Parts and then on your

    \int VdU

    Integral let

    p=x^2 + 9 --> x^2 = p -9

    This should work to get this to a nice integral for yeah
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by AllanCuz View Post
    \int 3(x^2+9)^{10}dx

    This is not a candidate for regular subsitution. There is no x in front of our bracket. In order to use regular substution whatever is in front of the bracket must be 1 order less then the term(s) in the bracket. In this case we would need an X.

    You cannt evaluate via this method.

    What you can do is integrate by Parts and then on your

    \int VdU

    Integral let

    p=x^2 + 9 --> x^2 = p -9

    This should work to get this to a nice integral for yeah
    Are you sure..?

    Obviously we must set u (as in u from uv - int vu') as (x^2 + 9)^{10} since if we knew how to integrate that we wouldn't need to use integration by parts...

    So we get...

    u = (x^2 + 9)^{10}

    u' = 20x(x^2 + 9)^9

    v = 3x

    Hence we get...

    3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx ...

    Setting x^2 = p-9 gives...

    x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp

    Hence ignoring the uv part the integral part becomes...

    \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp

    Is this easy? I can't see an easy way..?

    I'm actually stumped by this question as well...

    Perhaps we need a double substitution or even a trig sub in the above integral...
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Deadstar View Post
    Are you sure..?

    Obviously we must set u (as in u from uv - int vu') as (x^2 + 9)^{10} since if we knew how to integrate that we wouldn't need to use partial integration...

    So we get...

    u = (x^2 + 9)^{10}

    u' = 20x(x^2 + 9)^9

    v = 3x

    Hence we get...

    3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx ...

    Setting x^2 = p-9 gives...

    x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp

    Hence ignoring the uv part the integral part becomes...

    \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp

    Is this easy? I can't see an easy way..?

    I'm actually stumped by this question as well...

    Perhaps we need a double substitution or even a trig sub in the above integral...
    Yeah I didn't fully write it out on paper so I thought it might simpify LOL
    You could do trig in your integral or we could find a reduction of order, seeing as this will continue to reduce (in terms of the variables). But yeah, i don't see this as being overly easy at all...

    Edit--What you did was essentially the method of Integration by Parts. When someone says U subsitution I usually take it to mean U=Inside the bracket, which in this case wasn't possible.
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Deadstar View Post
    Are you sure..?

    Obviously we must set u (as in u from uv - int vu') as (x^2 + 9)^{10} since if we knew how to integrate that we wouldn't need to use integration by parts...

    So we get...

    u = (x^2 + 9)^{10}

    u' = 20x(x^2 + 9)^9

    v = 3x

    Hence we get...

    3x(x^2 + 9)^{10} - \int 60x^2(x^2 + 9)^9 dx ...

    Setting x^2 = p-9 gives...

    x = \sqrt{p-9}, dx = \frac{1}{2} (p-9)^{-1/2} dp

    Hence ignoring the uv part the integral part becomes...

    \int \frac{60(p-9)p^9}{2 \sqrt{p-9}} dp

    Is this easy? I can't see an easy way..?

    I'm actually stumped by this question as well...

    Perhaps we need a double substitution or even a trig sub in the above integral...
    Yes. This question is not as easy and as small as it looks. Here's Wolfram Alpha for you:

    integrate(3*(x^2+9)& #x5e;10) - Wolfram|Alpha
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  6. #6
    Super Member Deadstar's Avatar
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    Quote Originally Posted by harish21 View Post
    Yes. This question is not as easy and as small as it looks. Here's Wolfram Alpha for you:

    integrate(3*(x^2+9)& #x5e;10) - Wolfram|Alpha
    I think it's easier than that. That is just the binomial expansion being integrated I think. I.e. (I think this is the right expansion...)
    <br />
3 \int (x^2 + 9)^{10} dx = 3 \int \sum_{j=0}^{10} x^{2j} 9^{10-j} {10 \choose j} dx

    So we could just integrate that to get...

    3\sum_{j=0}^{10} \frac{x^{2j+1} 9^{10-j}}{2j+1} {10 \choose j}

    But I believe there will be a more elegant solution...
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  7. #7
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Deadstar View Post
    I think it's easier than that. That is just the binomial expansion being integrated I think. I.e. (I think this is the right expansion...)
    <br />
3 \int (x^2 + 9)^{10} dx = 3 \int \sum_{j=0}^{10} x^{2j} 9^{10-j} {10 \choose j} dx

    So we could just integrate that to get...

    3\sum_{j=0}^{10} \frac{x^{2j+1} 9^{10-j}}{2j+1} {10 \choose j}

    But I believe there will be a more elegant solution...
    I talked with my professor and he suggested hyperbolic substitution but the problem is when you get to the reduction formula part you need to compute all the way from I21 --> I1 and this is almost as long as if we were to expand the binomial!

    In short, expand the binomial lol.
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