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Math Help - Integrate sqrt(1 +a^2*x^2)dx

  1. #1
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    Integrate sqrt(1 +a^2*x^2)dx

    Integrate

    sqrt(1 +a^2*x^2) dx

    Thanks
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  2. #2
    GAMMA Mathematics
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    You want to do a trig substitution so that the square root term will disappear. Have you tried subbing tangent, secant, cotangent, or cosecant in for x?
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  3. #3
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    i have been trying x = a*sinh u
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by j-lee00 View Post
    i have been trying x = a*sinh u
    Try x=\frac{tan(t)}{a}

    dx=\frac{sec^2(t)}{a}dt

    \int\sqrt{1+tan^2(t)}\frac{sec^2(t)}{a}dt

    \frac{1}{a}*\int sec^3(t)dt

    \frac{{sec(t)tan(t)}^3}{a}

    t = tanh(ax)

    \frac{{sec(tanh(ax))tan(tanh(ax))}^3}{a}

    \frac{{sec(tanh(ax))(ax))}^3}{a}
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  5. #5
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by j-lee00 View Post
    i have been trying x = a*sinh u
    x=\frac{\sinh{u}}{a} should work.
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  6. #6
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    Hello, j-lee00!

    Integrate: . \int \sqrt{1 +a^2x^2}\,dx

    Let: ax \:=\:\tan\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{a}\sec^2\!\theta\,d\theta
    Then: . \sqrt{1+a^2x^2} \:=\:\sec\theta

    Substitute: . \int\sec\theta\left(\frac{1}{a}\sec^2\!\theta\,d\t  heta\right) \;=\;\frac{1}{a}\int\sec^3\!\theta\,d\theta


    This integral requires Integration By Parts twice, or a standard formula.

    . . \frac{1}{2a}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C



    Back-substitute . . .

    \text{Since }\tan\theta \:=\:\frac{ax}{1} \:=\:\frac{opp}{adj},\:\text{ then: }\:hyp \:=\:\sqrt{1+a^2x^2} \quad\Rightarrow\quad \sec\theta \:=\:\sqrt{1+a^2x^2}


    Answer: . \frac{1}{2a}\bigg[ax\sqrt{1+a^2x^2} \:+ \:\ln\left|ax + \sqrt{1+a^2x^2}\right|\bigg] + C

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  7. #7
    GAMMA Mathematics
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    J Lee,

    I didn't work with the geometry, but Soroban basically completed what I started for you.

    -Colby
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  8. #8
    Senior Member JaneBennet's Avatar
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    J Lee was trying to use a sinh substitution; I think the integration is easier with this substitution.

    x=\frac{\sinh{u}}{a}\ \Rightarrow\ \mathrm{d}x=\frac{\cosh{u}}{a}\,\mathrm{d}u

    \begin{array}{rcl}<br />
\therefore\ \int{\sqrt{1+a^2x^2}}\,\mathrm{d}x &=& \frac{1}{a}\int{\sqrt{1+\sinh^2{u}}\cdot\cosh{u}}\  ,\mathrm{d}x\\\\<br />
{} &=& \frac{1}{a}\int{\cosh^2{u}}\,\mathrm{d}x\\\\<br />
{} &=& \frac{1}{2a}\int{(\cosh{2u}+1)}\,\mathrm{d}x\\\\<br />
{} &=& \frac{1}{2a}\,(\frac{1}{2}\sinh{2u}+u)+C\\\\<br />
{} &=& \frac{1}{2a}\,(\sinh{u}\cosh{u}+u)+C\\\\<br />
\end{array}

    Now you can substitute the x back (noting that \sinh^{-1}{ax}=\ln{(ax+\sqrt{1+a^2x^2})}).
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  9. #9
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    thanks for all your assistances
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