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Math Help - integral question

  1. #1
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    integral question

    Hi I am working on a integral involving u substiution and inverse trig forms.

     <br />
Evaluate:~~\int\frac{2}{(2x+3)\sqrt{4x^2+12x+5}}dx =<br />

     <br />
\int\frac{2}{(2x+3)\sqrt{\frac{x^2+3x+\frac{5}{4}}  {\frac{1}{4}}}}dx<br />
=

    Complete the square of the numberator of the radical:
     <br />
\int\frac{2}{(2x+3)\sqrt{\frac{(x+\frac{3}{4})^2-1}{\frac{1}{4}}}}dx<br />
=

    Partially expand the square-completed numerator:
    \int\frac{2}{(2x+3)\sqrt{\frac{(x^2+3x+\frac{9}{4}  )-1}{\frac{1}{4}}}}dx<br />
=

    Simplify the expression in the radical by elimating the denominator:
    \int\frac{2}{(2x+3)\sqrt{(4x^2+12x+9)-4}}dx

    Factor the portion in parenthesis in the radical:
    \int\frac{2}{(2x+3)\sqrt{(2x+3)^2-4}}dx

    Let u = 2x + 3
    then du = 2dx.

    So now we have:
     <br />
\int\frac{1}{u\sqrt{u^2-4}}du<br />

    Notice this is the general arcsec integral formula, and so we have:

     <br />
\frac{1}{2}sec^{-1}\left(\frac{|u|}{2}\right) + C<br />

    So substituting back in for u, we have:
     <br />
\frac{1}{2}sec^{-1}\left(\frac{|2x+3|}{2}\right)+C<br />

    I was pretty confident this was right but when I checked it using mathematica I got:

     <br />
-\frac{1}{2}tan^{-1}\frac{2}{\sqrt{(2x+3)^2-4}}, which I checked as not being equivalent to my answer. Where might I have gone wrong?
    Thanks for your help,
    James
    Last edited by james121515; January 5th 2010 at 05:39 PM. Reason: typo
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by james121515 View Post
    Hi I am working on a integral involving u substiution and inverse trig forms.

     <br />
Evaluate:~~\int\frac{2}{(2x+3)\sqrt{4x^2+12x+5}}dx =<br />

     <br />
\int\frac{2}{(2x+3)\sqrt{\frac{x^2+3x+\frac{5}{4}}  {\frac{1}{4}}}}dx<br />
=

    Complete the square of the numberator of the radical:
     <br />
\int\frac{2}{(2x+3)\sqrt{\frac{(x+\frac{3}{4})^2-1}{\frac{1}{4}}}}dx<br />
=

    Partially expand the square-completed numerator:
    \int\frac{2}{(2x+3)\sqrt{\frac{(x^2+3x+\frac{9}{4}  )-1}{\frac{1}{4}}}}dx<br />
=

    Simplify the expression in the radical by elimating the denominator:
    \int\frac{2}{(2x+3)\sqrt{(4x^2+12x+9)-4}}dx

    Factor the portion in parenthesis in the radical:
    \int\frac{2}{(2x+3)\sqrt{(2x+3)^2-4}}dx

    Let u = 2x + 3
    then du = 2dx.

    So now we have:
     <br />
\int\frac{1}{u\sqrt{u^2-4}}du<br />

    Notice this is the general arcsec integral formula, and so we have:

     <br />
\frac{1}{2}sec^{-1}\left(\frac{|u|}{2}\right) + C<br />

    So substituting back in for u, we have:
     <br />
\frac{1}{2}sec^{-1}\left(\frac{|2x+3|}{2}\right)+C<br />

    I was pretty confident this was right but when I checked it using mathematica I got:

     <br />
-\frac{1}{2}tan^{-1}\frac{2}{\sqrt{(2x+3)^2-4}}, which I checked as not being equivalent to my answer. Where might I have gone wrong?
    Thanks for your help,
    James
    you made completing the square a lot harder than it should be, but what you did is fine.

    i wasn't sure about the absolute values in your answer though... one of my calc texts has them, the other doesn't
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