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Thread: Integral.

  1. #1
    Ted
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    Integral.

    Hello.
    evaluate:

    $\displaystyle \int \sqrt{x^2+16} dx$.
    Without using Trig/Hyperbolic Substitutions.
    And you must use the substitution method only.
    I believed, its impossible to solve it.
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  2. #2
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    The way I was taught to answer integrals:
    $\displaystyle \sqrt{x^2+16} dx$

    $\displaystyle
    =(x^2+16)^\frac{1}{2}dx
    $

    $\displaystyle
    u = x^2+16 =>
    du = 2x
    $

    $\displaystyle
    =\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]
    $
    $\displaystyle
    =\frac{1}{2}*
    \frac{u^{3/2}}{\frac{3}{2}}
    $

    Answer:
    $\displaystyle
    = \frac{1}{3}(x^2+16)^{3/2}
    $
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  3. #3
    MHF Contributor

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    Quote Originally Posted by lil_cookie View Post
    The way I was taught to answer integrals:
    $\displaystyle \sqrt{x^2+16} dx$

    $\displaystyle
    =(x^2+16)^\frac{1}{2}dx
    $

    $\displaystyle
    u = x^2+16 =>
    du = 2x
    $

    $\displaystyle
    =\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]
    $
    Unfortunately, you cannot just put that "x" into the integrand!

    $\displaystyle
    =\frac{1}{2}*
    \frac{u^{3/2}}{\frac{3}{2}}
    $

    Answer:
    $\displaystyle
    = \frac{1}{3}(x^2+16)^{3/2}
    $
    No, the derivative of $\displaystyle \frac{1}{3}(x^2+ 16)^{3/2}$ is
    $\displaystyle (3/2)(1/3)(x^2+ 16)^{1/2} (2x)= x(x^2+16)^{1/2}$, not just $\displaystyle (x^2+ 16)^{1/2}$

    I doubt that you were taught that way!

    Ted, I don't believe there is any way to do that other than either the trig or hyperbolic substitutions.

    Of course, you could always look it up in a table of integrals. Does that count?
    http://www.integral-table.com/
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  4. #4
    Math Engineering Student
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    Quote Originally Posted by Ted View Post
    Hello.
    evaluate:

    $\displaystyle \int \sqrt{x^2+16} dx$.
    Without using Trig/Hyperbolic Substitutions.
    put $\displaystyle x=4u$ and the integral becomes $\displaystyle 16\int\sqrt{u^2+1}\,du.$

    put $\displaystyle t=\sqrt{u^2+1}+u$ then $\displaystyle \frac1t=\sqrt{u^2+1}-u,$ thus $\displaystyle \frac12\left(t+\frac1t\right)=\sqrt{u^2+1}$ and $\displaystyle \frac12\left(t-\frac1t\right)=u$ so i think you can finish it.
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