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Math Help - Integration By Parts Problem

  1. #1
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    Integration By Parts Problem

    Find the integral of (x cos(2x)) dx, using integration by parts.

    I used x as dv/dx and cos(2x) as u.

    So v = (v^2/2) and (du/dx) = -2sin(2x)

    I am having problems using the formula:

    I = uv - Integral ( u' v ) dx. Can anyone help me with this as I can't get the correct answer.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by haku View Post
    Find the integral of (x cos(2x)) dx, using integration by parts.

    I used x as dv/dx and cos(2x) as u.

    So v = (v^2/2) and (du/dx) = -2sin(2x)

    I am having problems using the formula:

    I = uv - Integral ( u' v ) dx. Can anyone help me with this as I can't get the correct answer.
    why would you use x as dv??

    let u = x and dv = cos(2x)
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  3. #3
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    I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by haku View Post
    I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.
    do not only consider what is easier to integrate, consider what you'll get when you are done. if you integrate x you will get x^2 times a trig function to integrate as opposed to just x times a trig function. you'd make your problem worst! if you differentiate x though, you will only have a trig function to integrate
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  5. #5
    Super Member wingless's Avatar
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    Quote Originally Posted by haku View Post
    I thought it would be easier to integrate the x. I'll have a go with u=x, I'll post if I have any problems. Thanks.
    Yes, it's easier to integrate it but it won't bring you anywhere. But if we differentiate x, it'll be 1 and if we differentiate 1, we get 0. This means you will surely get a result using integration by parts 2 times. Learn tabular integration, it makes easier to use multiple partial integrations.

    If you have troubles integrating this, you can substitute t=2x before starting.

    \int x \cos 2x ~dx
    t=2x
    dt = 2dx
    x=\frac{t}{2}

    So it becomes \frac{1}{4} \int t \cos t ~dt which is easier to integrate.
    Last edited by wingless; February 12th 2008 at 02:32 PM.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wingless View Post
    Yes, it's easier to integrate it but it won't bring you anywhere. But if we differentiate x, it'll be 1 and if we differentiate 1, we get 0. This means you will surely get a result using integration by parts 2 times. Learn tabular integration, it makes easier to use multiple partial integrations.

    If you have troubles integrating this, you can substitute u=2x before starting.

    \int x \cos 2x ~dx
    u=2x
    du = 2dx
    x=\frac{u}{2}

    So it becomes \frac{1}{4} \int u \cos u ~du which is easier to integrate.
    you should use t or something for the substitution, since the poster uses u for the integration by parts formula. there may be confusion.
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  7. #7
    Super Member wingless's Avatar
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    Quote Originally Posted by Jhevon View Post
    you should use t or something for the substitution, since the poster uses u for the integration by parts formula. there may be confusion.
    Haha, you're right. Edited
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  8. #8
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    So for this example:

    Integral (Limits 0 to 1) (2x-1) e^(3x+2).

    I set (dv/dx)=(2x-1) so, v=(((2x-1)^2)/2)

    and u=e^(3x+2) so, u'=3e^(3x+2).

    Am I correct so far. Then using the formula:

    e^(3x+2).(((2x-1)^2)/2) - (Integral of) (3e^(3x+2)).(((2x-1)^2)/2).

    Would you use integration by parts again for the second part? I can't seem to do this. Please help.
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  9. #9
    Super Member wingless's Avatar
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    \int^{1}_{0} (2x-1)e^{3x+2}~dx

    First simplify the function. This will make it really easy to integrate.

    \int^{1}_{0} (2x-1)e^{3x}e^{2}~dx

    e^{2}\int^{1}_{0} (2x-1)e^{3x}~dx

    Let u=2x-1, dv = e^{3x} and integrate it by parts two times.
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  10. #10
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    First, keep simplifying.

    \int{(2x-1)e^{3x}}\;dx\;=\;2\int{xe^{3x}}\;dx\;-\;\int{e^{3x}}\;dx\;=\;

    Second, why twice?

    \int{xe^{3x}}\;dx\;=\;\int{x}\;d\left(\frac{e^{3x}  }{3}\right)\;=\;\frac{xe^{3x}}{3}\;-\;\int{\frac{e^{3x}}{3}}\;dx
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