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Math Help - trigonometric and hyperbolic substitution

  1. #1
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    trigonometric and hyperbolic substitution

    Hi all,

    My apologises to angel.white, i completely typed the incorrect question and i would like to thank Isomorphism

    Using an appropriate trigonometric or hyperbolic substitution, find the indefinite integrals of
    sqrt( 1 + 4x^2)

    Thanks in advance,
    ArTiCk
    Last edited by ArTiCK; June 2nd 2008 at 04:52 AM.
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  2. #2
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I have a feeling the answer to the following question is incorrect:

    Using an appropriate trigonometric or hyperbolic substitution, find the indefinite integrals of
    <br />
\int{{1 + 4x^2}} dx<br />

    I got 1/4 (x sqrt( 1 + 4x^2) + 1/2arcsinh2x) + c...

    however the answer says: 1/4(arcsinh(2x) + 2x*sqrt(1 + 4x^2)) + C

    Thanks in advance,
    ArTiCk
    Well, if you don't use a trig/hyperbolic substitution, you get \int\left(1+4x^2\right)~dx = x+\frac{4x^3}3+C

    So unless I really missed something, both answers must be wrong.
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  3. #3
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    Hi,

    It says 'using a hyperbolic or trigonometric substitution'

    ArTiCk
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by ArTiCK View Post
    Hi,

    It says 'using a hyperbolic or trigonometric substitution'

    ArTiCk
    \int\left(1+4x^2\right)~dx

    -------------------------
    substitution:
    sin(a) = x

    cos(a) ~da = ~dx
    -------------------------
    =\int cos(a)\left(1+4sin^2(a)\right)~da

    =\int cos(a) ~da + 4 \int cos(a)~sin^2(a)~da

    =sin(a) + \frac 43sin^3(a)+C

    anti-substitute

    =x +\frac 43x^3+C

    Regardless of what method you use, you should arrive at the same conclusion. Are you sure you copied the problem correctly?
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  5. #5
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    Quote Originally Posted by angel.white View Post
    \int\left(1+4x^2\right)~dx

    -------------------------
    substitution:
    sin(a) = x

    cos(a) ~da = ~dx
    -------------------------
    =\int cos(a)\left(1+4sin^2(a)\right)~da

    =\int cos(a) ~da + 4 \int cos(a)~sin^2(a)~da

    =sin(a) + \frac 43sin^3(a)+C

    anti-substitute

    =x +\frac 43x^3+C

    Regardless of what method you use, you should arrive at the same conclusion. Are you sure you copied the problem correctly?
    Yes i am totally sure that i copied it correctly, because i have already done some and have the correct answer.... besides this one

    Thanks for trying though
    ArTiCk
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  6. #6
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    Quote Originally Posted by ArTiCK View Post
    Hi,

    It says 'using a hyperbolic or trigonometric substitution'

    ArTiCk

    If it said so, I am nearly convinced that you must have missed a square root sign. Check it again carefully. Polynomial integration does not need 'a hyperbolic or trigonometric substitution'.

    Most likely, the question was:

    \int \sqrt{1+4x^2} \, dx
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  7. #7
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    Quote Originally Posted by Isomorphism View Post
    If it said so, I am nearly convinced that you must have missed a square root sign. Check it again carefully. Polynomial integration does not need 'a hyperbolic or trigonometric substitution'.

    Most likely, the question was:

    \int \sqrt{1+4x^2} \, dx
    I apologize to angel.white and thank you isomorphism for picking up the mistake. The correct equation is in the first post.

    Thank you once again,
    ArTiCk
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  8. #8
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    Quote Originally Posted by ArTiCK View Post
    I apologize to angel.white and thank you isomorphism for picking up the mistake. The correct equation is in the first post.

    Thank you once again,
    ArTiCk
    Substitute x = \frac{1}{2} \sinh t \Rightarrow dx = \frac{1}{2} \cosh t\, dt.

    Then the integral becomes \frac{1}{2} \int \cosh^2 t \, dt .....

    Use Osborn's Rule to get the appropriate 'double angle' formula to deal with this .......
    Last edited by mr fantastic; June 2nd 2008 at 05:30 AM. Reason: Lost a 'dt' .....
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    Substitute x = \frac{1}{2} \sinh t \Rightarrow dx = \frac{1}{2} \cosh t.

    Then the integral becomes \frac{1}{2} \int \cosh^2 t \, dt .....

    Use Osborn's Rule to get the appropriate 'double angle' formula to deal with this .......
    For checking purposes:

    x = \frac{1}{2} \sinh t \Rightarrow t = \text{arcsinh} \, (2x). Therefore:

    \frac{1}{4} t \rightarrow \frac{1}{4} \text{arcsinh} \, (2x).

    \frac{1}{8} \sinh (2t) = \frac{1}{4} \sinh t \cosh t \rightarrow \frac{1}{4} (2x) \sqrt{1 + 4x^2} = \frac{1}{2} \, x \, \sqrt{1 + 4x^2}.
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