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Math Help - Integral - trigonometric substitution?

  1. #1
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    Integral - trigonometric substitution?

    Solve the following problem using trig substitution:

    \int{\frac{x}{\sqrt{x^2+x+1}}}
    Last edited by mr fantastic; December 17th 2010 at 11:45 AM. Reason: Title.
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  2. #2
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    I think you forgot a dx.

    \frac{x}{\sqrt{x^2+x+1}} = \frac{x}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}

    Let y = x + \frac{1}{2}
    \frac{y - \frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}} = \frac{y}{\sqrt{y^2+\frac{3}{4}}} - \frac{\frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}}

    Now, its split into one simple integral \frac{y}{\sqrt{y^2+\frac{3}{4}}} and a trig sub \frac{\frac{1}{2}}{\sqrt{y^2+\frac{3}{4}}}.

    Use y^2 = \frac{3}{4}tan^2(\theta)
    Last edited by snowtea; December 17th 2010 at 08:17 AM.
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  3. #3
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    i did forget the dx, sorry.
    but wouldn't (x+\frac{1}{2})^2+\frac{1}{2}=x^2+x+\frac{3}{4} and not x^2+x+1
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  4. #4
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    Yes, you are absolutely right
    Let me fix that.
    But you get the idea.
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  5. #5
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    I solved this integral in here, and here, and here.
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  6. #6
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    thank you, but is there any other way of doing this problem without setting y= to some number?
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  7. #7
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    Hello, drewbear!

    This problem is quite awful!


    \displaystyle \text{Solve using trig substitution: }\:\int \frac{x\,dx}{\sqrt{x^2+x+1}}

    Under the radical, complete the square:

    . . x^2 + x + 1 \;=\;x^2 + x + \frac{1}{4} + \frac{3}{4} \;=\;\left(x+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2

    The integral becomes: . \displaystyle \int \frac{x\,dx} {\sqrt{(x + \frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2}}


    Let x + \frac{1}{2} \:=\:\frac{\sqrt{3}}{2}\tan\theta \quad\Rightarrow\quad x \:=\:\frac{\sqrt{3}}{2}\tan\theta - \frac{1}{2} \quad\Rightarrow\quad dx \:=\:\frac{\sqrt{3}}{2}\sec^2\!\theta\,d\theta


    Substitute: . \displaystyle \int\frac{\left(\frac{\sqrt{3}}{2}\tan\theta - \frac{1}{2}\right)<br />
\left(\frac{\sqrt{3}}{2}\sec^2\theta\,d\theta)}{\f  rac{\sqrt{3}}{2}\sec\theta}

    . . . . . \displaystyle =\; \tfrac{1}{2}\int(\sqrt{3}\tan\theta - 1)\,\sec\theta\,d\theta

    . . . . . \displaystyle =\; \tfrac{1}{2}\int (\sqrt{3}\sec\theta\tan\theta - \sec\theta)\,d\theta

    . . . . . =\;\frac{1}{2}\bigg[\sqrt{3}\sec\theta + \ln|\sec\theta + \tan\theta|\bigg] + C


    Back-substitute: . \displaystyle \tan\thet \:=\:\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}} \;=\;\frac{2x+1}{\sqrt{3}}

    . . \displaystyle \sec^2\!\theta \:=\:\tan^2\!\theta + 1 \:=\:\left(\frac{2x+1}{\sqrt{3}}\right)^2 + 1 \;=\; \frac{4(x^2+x+1)}{3}

    . . \sec\theta \;=\;\frac{2}{\sqrt{3}}\sqrt{x^2+x+1}


    We have: . \frac{1}{2}\left[\sqrt{3}\cdot\frac{2}{\sqrt{3}}\sqrt{x^2+x+1} - \ln\left|\frac{2\sqrt{x^2+x+1}}{\sqrt{3}} + \frac{2x+1}{\sqrt{3}}\right|\, \right] + C

    . . . . . . =\;\sqrt{x^2+x+1} - \frac{1}{2}\ln\left|\frac{2\sqrt{x^2+x+1} + 2x + 1}{\sqrt{3}}\right| + C


    This can be simplified further, but I won't bother . . .
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  8. #8
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    Quote Originally Posted by drewbear View Post
    Solve the following problem using trig substitution:

    \int{\frac{x}{\sqrt{x^2+x+1}}}
    Maybe an alcoholic substitution would work well here...

    \displaystyle \int{\frac{x}{\sqrt{x^2+x+1}}\,dx} = \int{\frac{x}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}}}\,dx}

    Make the substitution \displaystyle x + \frac{1}{2} = \frac{\sqrt{3}}{2}\sinh{t} so that \displaystyle dx = \frac{\sqrt{3}}{2}\cosh{t}\,dt and the integral becomes

    \displaystyle \int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\sqrt{\frac{3}{4}\sinh^2{t} + \frac{3}{4}}}\left(\frac{\sqrt{3}}{2}\cosh{t}\righ  t)\,dt}

    \displaystyle =\int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\sqrt{\frac{3}{4}}\sqrt{\sinh^2{t} + 1}}\left(\frac{\sqrt{3}}{2}\cosh{t}\right)\,dt}

    \displaystyle =\int{\frac{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}}{\frac{\sqrt{3}}{2}\cosh{t}}\left(\fra  c{\sqrt{3}}{2}\cosh{t}\right)\,dt}

    \displaystyle =\int{\frac{\sqrt{3}}{2}\sinh{t} - \frac{1}{2}\,dt}

    \displaystyle = \frac{\sqrt{3}}{2}\cosh{t} - \frac{t}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\sqrt{\sinh^2{t} + 1} - \frac{t}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\sqrt{\left(\frac{2x+1}{\sqrt{3}  }\right)^2 + 1}- \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4x^2 + 4x + 1}{3} + \frac{3}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4x^2 + 4x + 4}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\sqrt{\frac{4(x^2+x+1)}{3}} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C

    \displaystyle = \frac{\sqrt{3}}{2}\frac{2}{\sqrt{3}}\sqrt{x^2+x+1} - \frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C

    \displaystyle = \sqrt{x^2+x+1} -\frac{\sinh^{-1}{\left(\frac{2x+1}{\sqrt{3}}\right)}}{2} + C
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