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Math Help - integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

  1. #1
    s3a
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    integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    I forgot how to do these kinds of problems. Could someone please show me how to find this integral (http://www.wolframalpha.com/input/?_=1322280641770&i=integrate+sqrt%2872^2t^4+%2b+48 ^2t^2%2b16^2%29&fp=1&incTime=true) ? Wolfram Alpha doesn't show me the steps for this; it just gives a final answer.

    Any help would be greatly appreciated!
    Thanks in advance!
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Quote Originally Posted by s3a View Post
    I forgot how to do these kinds of problems. Could someone please show me how to find this integral (http://www.wolframalpha.com/input/?_=1322280641770&i=integrate+sqrt%2872^2t^4+%2b+48 ^2t^2%2b16^2%29&fp=1&incTime=true) ? Wolfram Alpha doesn't show me the steps for this; it just gives a final answer.

    Any help would be greatly appreciated!
    Thanks in advance!
    I would start by completing the square...
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    The expression 72^2 t^4+48^2 t+16^2 is of the form a^2+b^2+2ab.

    72^2 t^4+48^2 t+16^2 = (72t^2+16)^2

    So \sqrt{(72t^2+16)^2}=?
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    s3a
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

    For example:
    x^2 - 2x + y^2 = 0
    (x - 2)^2 + y^2 = 0 + 1

    Did you mean what sbhatnagar said? If not, could you show me your way too please?
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Quote Originally Posted by s3a View Post
    Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

    For example:
    x^2 - 2x + y^2 = 0
    (x - 2)^2 + y^2 = 0 + 1

    Did you mean what sbhatnagar said? If not, could you show me your way too please?
    If you are integrating at this level you ought to know how to complete the square on something that has the form 72^2w^2 + 48^2w + 16^2:

    complete the square 72^2w^2 + 48^2w + 16^2 - Wolfram|Alpha

    Obviously Prove It meant what sbhatnagar said. How else would you understand "complete the square"?
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Quote Originally Posted by s3a View Post
    Prove it: I don't see how to complete the square for this. I do know how to complete the square though generally.

    For example:
    x^2 - 2x + y^2 = 0
    (x - 1)^2 + y^2 = 0 + 1

    Did you mean what sbhatnagar said? If not, could you show me your way too please?
    f(t) = 72^2t^4 + 48^2t^2 + 16^2

    Let u = t^2 and a quadratic will suddenly appear: 72^2u^2 + 48^2u + 16^2. Complete the square as you would normally:

    72^2\left(u^2 + \left(\dfrac{48}{72}\right)^2u + \left(\dfrac{16}{72}\right)^2\right) = 72^2\left(u^2 + \left(\dfrac{2}{3}\right)^2u + \left(\dfrac{2}{9}\right)^2\right)
    Last edited by e^(i*pi); November 26th 2011 at 11:48 AM. Reason: leaving OP something to do
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    72^2t^4 + 48^2t^2 + 16^2 =

    (2^6 \cdot 3^4)t^4 + (3^2 \cdot 2^8)t^2 + 2^8 =

    2^6[3^4t^4 + (3^2 \cdot 2^2) t^2 + 2^2] =

    2^6(3^2t^2 + 2)^2 =

    64(9t^2+2)^2
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Hello, s3a;697799!

    \int \sqrt{ 72^2t^4+ 48^2t^2 + 16^2}\,dt
    It is unfortunate that "complete the square" was mentioned.


    Factor the given polynomial:

    . . 8^2\left(9^2t^4 + 6^2t^2 + 2^2\right) \:=\:64(81t^4 + 36t^2 + 4) \:=\:64(9t^2+2)^2


    The integral becomes:
    . . \int\sqrt{64(9t^2+2)^2}\,dt \;=\;\int 8(9t^2+2)\,dt \;=\;8\int(9t^2+2)\,dt

    Got it?


    Edit: Too slow . . . again!
    .
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    Re: integrating sqrt(72^2t^4 + 48^2t^2+16^2) dt

    Quote Originally Posted by Soroban View Post
    Hello, s3a;697799!

    It is unfortunate that "complete the square" was mentioned.


    Factor the given polynomial:

    . . 8^2\left(9^2t^4 + 6^2t^2 + 2^2\right) \:=\:64(81t^4 + 36t^2 + 4) \:=\:64(9t^2+2)^2


    The integral becomes:
    . . \int\sqrt{64(9t^2+2)^2}\,dt \;=\;\int 8(9t^2+2)\,dt \;=\;8\int(9t^2+2)\,dt

    Got it?


    Edit: Too slow . . . again!
    .
    For anyone who wants a general method which will work for similar problems, completing the square is the best method. It is also a valid method in this case.
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