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Math Help - integrate by substitution

  1. #1
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    integrate by substitution

    how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

    my working:

    let 6+x+ x^2 = (x+2)^2 (z^2)

    then dx = -10z / ( z^2 +1) ^(2) dz

    substituting that into the question, i got

    integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))



    however the answer should have been arcsin ( (2x-1)/5)).

    is there something wrong with my substitution?
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  2. #2
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    Have you considered Completing the Square and then a Trigonometric substitution?
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  3. #3
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    Quote Originally Posted by alexandrabel90 View Post
    how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

    my working:

    let 6+x+ x^2 = (x+2)^2 (z^2)

    then dx = -10z / ( z^2 +1) ^(2) dz

    substituting that into the question, i got

    integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))


    however the answer should have been arcsin ( (2x-1)/5)).

    is there something wrong with my substitution?
    \int{\frac{1}{\sqrt{x^2 + x + 6}}\,dx} = \int{\frac{1}{\sqrt{x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 6}}\,dx}

     = \int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{23}{4}}}\,dx}.


    Now use the substitution \frac{\sqrt{23}}{2}\tan{\theta} = x + \frac{1}{2}

    So \frac{dx}{d\theta}= \frac{\sqrt{23}}{2}\sec^2{\theta}

    dx = \frac{\sqrt{23}}{2}\sec^2{\theta}\,d\theta.

    So the integral becomes

    \int{\frac{1}{\sqrt{\left(\frac{\sqrt{23}}{2}\tan{  \theta}\right)^2 + \frac{23}{4}}}\,\frac{\sqrt{23}}{2}\sec^2{\theta}\  ,d\theta}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt  {23}}{2}(\tan^2{\theta} + 1)}}\,\sec^2{\theta}\,d\theta}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt  {23}}{2}(\sec^2{\theta})}}\,\sec^2{\theta}\,d\thet  a}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt  {23}}{2}}\sec{\theta}}\,\sec^2{\theta}\,d\theta}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{\sqrt  {23}}{2}}\sec{\theta}}\,\sec^2{\theta}\,d\theta}

    = \frac{\sqrt{23}}{2}\,\sqrt{\frac{2}{\sqrt{23}}}\in  t{\sec{\theta}\,d\theta}

     = \frac{\sqrt{23}}{2}\,\sqrt{\frac{2\sqrt{23}}{23}}\  int{\sec{\theta}\,d\theta}

     = \frac{\sqrt{2\sqrt{23}}}{2}\int{\sec{\theta}\,d\th  eta}

     = \frac{\sqrt{2\sqrt{23}}}{2}\ln{|\sec{\theta} + \tan{\theta}|}


    Now get this function back in terms of x.

    Remember that

    \frac{\sqrt{23}}{2}\tan{\theta} = x + \frac{1}{2}

    and \sec{\theta} = \sqrt{1 + \tan^2{\theta}}.


    P.S. I do not agree with the answer given. If they want the integral to contain arcsin, the integrand should have been of the form \frac{1}{\sqrt{a^2 - x^2}}, not \frac{1}{\sqrt{a^2 + x^2}} as yours was.

    Are you sure you copied the question down correctly?
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  4. #4
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    sorry the question should have been
    integrate 1/ ( 6+x- x^2 )^(1/2) instead.
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  5. #5
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    Quote Originally Posted by alexandrabel90 View Post
    how do you integrate 1/ ( 6+x+ x^2 )^(1/2)

    my working:

    let 6+x+ x^2 = (x+2)^2 (z^2)

    then dx = -10z / ( z^2 +1) ^(2) dz

    substituting that into the question, i got

    integrate -2 / (z^2+1) and hence the answer was -2 arctan ( ((3-x)/ (x+2))^(1/2))



    however the answer should have been arcsin ( (2x-1)/5)).

    is there something wrong with my substitution?
    Upon doing some further research, this question is much easier using a hyperbolic substitution, rather than a trigonometric.

    Your integral is

    \int{\frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \frac{23}{4}}}\,dx}.

    So you let x + \frac{1}{2} = \frac{\sqrt{23}}{2}\sinh{t}.

    Therefore \frac{dx}{dt} = \frac{\sqrt{23}}{2}\cosh{t}, and so dx = \frac{\sqrt{23}}{2}\cosh{t}\,dt.


    So the integral becomes

    \int{\frac{1}{\sqrt{\left(\frac{\sqrt{23}}{2}\sinh  {t}\right)^2 + \frac{23}{4}}}\,\frac{\sqrt{23}}{2}\cosh{t}\,dt}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\sqrt{\frac{23}{4  }(\sinh^2{t} + 1)}}\,\cosh{t}\,dt}

     = \frac{\sqrt{23}}{2}\int{\frac{1}{\frac{\sqrt{23}}{  2}\sqrt{\cosh^2{t}}}\,\cosh{t}\,dt}

     = \frac{\sqrt{23}}{2}\,\frac{2}{\sqrt{23}}\int{\frac  {1}{\cosh{t}}\,\cosh{t}\,dt}

     = \int{1\,dt}

     = t + C.


    Now since x + \frac{1}{2} = \frac{\sqrt{23}}{2}\sinh{t}

    \sinh{t} = \frac{2}{\sqrt{23}}\left(x + \frac{1}{2}\right)

    \sinh{t} = \frac{2x + 1}{\sqrt{23}}

    t = \sinh^{-1}{\left(\frac{2x + 1}{\sqrt{23}}\right)}.


    Therefore the integral is

    \sinh^{-1}{\left(\frac{2x + 1}{\sqrt{23}}\right)} + C.
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  6. #6
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    Quote Originally Posted by alexandrabel90 View Post
    sorry the question should have been
    integrate 1/ ( 6+x- x^2 )^(1/2) instead.
    Right.

    \int{\frac{1}{\sqrt{6 + x - x^2}}\,dx} = \int{\frac{1}{\sqrt{-(x^2 - x - 6)}}\,dx}

     = \int{\frac{1}{\sqrt{-\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 - 6\right]}}\,dx}

     = \int{\frac{1}{\sqrt{-\left[\left(x - \frac{1}{2}\right)^2 - \frac{25}{4}\right]}}\,dx}

     = \int{\frac{1}{\sqrt{\frac{25}{4} - \left(x - \frac{1}{2}\right)^2}}\,dx}.


    Now use the substitution x - \frac{1}{2} = \frac{5}{2}\sin{\theta}.
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  7. #7
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    instead of using completing the square, can i solve the question if i factorise
    6+x-x^2 = (3-x)(x+2) ?
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  8. #8
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    Quote Originally Posted by alexandrabel90 View Post
    instead of using completing the square, can i solve the question if i factorise
    6+x-x^2 = (3-x)(x+2) ?
    No. The square root in the denominator prevents you from being able to factorise and use partial fractions.

    Have a go using the trigonometric subsitution.
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