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Math Help - Trig sub intehral w/ completing the sq

  1. #1
    Junior Member symstar's Avatar
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    Trig sub intehral w/ completing the sq

    \int \frac{dx}{(x^2+2x+2)^2}
    u=x+1 & du=dx
    \int \frac{du}{(u^2+1)^2}

    What should I do at this point? I considered expanding the square, but that seems to make things more messy and it seems like something else should be going on. It also didn't seem to lead me to an answer either. Thanks!
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  2. #2
    Super Member 11rdc11's Avatar
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    After you complete the square

    let x + 1 =tan(u)


    dx = sec^2(u)~du

    tan^2(u) + 1 = sec^2(u)

    so your integral looks like this

    \int \frac{sec^2(u)}{sec^4(u)}~du = \int \frac{1}{sec^2(u)}~du = \int cos^2(u)

    From here use half angle formula

    cos^2(u) = \frac{1+cos(2u)}{2}

    and you get

    \int \frac{1 + cos(2u)}{2}~du

    Now that should be an easier integral to evaluate and back substitute for u
    Last edited by 11rdc11; September 9th 2008 at 02:02 PM.
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  3. #3
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     \int {\frac{1}<br />
{{\left( {x^2  + 2x + 2} \right)^2 }}} dx = \int {\frac{1}<br />
{{\left[ {\left( {x + 1} \right)^2  + 1} \right]^2 }}}  \hfill \\

      {\text{Let us substitute, }}x + 1 = \tan \theta  \Rightarrow \theta  = \tan ^{ - 1} \left( {x + 1} \right) \hfill \\

    {\text{so that,  }}xdx = \sec ^2 \theta d\theta  \hfill \\

     {\text{Also, }}\tan \theta  = \frac{{x + 1}}<br />
{1} \Rightarrow \sin \theta  = \frac{{x + 1}}<br />
{{\left( {x + 1} \right)^2  + 1}} = \frac{{x + 1}}<br />
{{x^2  + 2x + 2}} \hfill \\

     {\text{and also, }} \Rightarrow \cos \theta  = \frac{1}<br />
{{\left( {x + 1} \right)^2  + 1}} = \frac{1}<br />
{{x^2  + 2x + 2}} \hfill \\

      {\text{Now see, }} = \int {\frac{1}<br />
{{\left( {\tan ^2 \theta  + 1} \right)^2 }}} \sec ^2 \theta d\theta  = \int {\frac{{\sec ^2 \theta }}<br />
{{\left( {\sec ^2 \theta } \right)^2 }}} d\theta  \hfill \\

       = \int {\cos ^2 \theta d\theta }  \hfill \\

       = \int {\frac{{1 + \cos 2\theta }}<br />
{2}} d\theta  \hfill \\

       = \frac{1}<br />
{2}\left( {\theta  + \sin 2\theta } \right) + c \hfill \\

       = \frac{1}<br />
{2}\left( {\theta  + 2\sin \theta \cos \theta } \right) + c \hfill \\

       = \frac{1}<br />
{2}\left[ {\tan ^{ - 1} \left( {x + 1} \right) + 2\left( {\frac{{x + 1}}<br />
{{x^2  + 2x + 2}}} \right)\left( {\frac{1}<br />
{{x^2  + 2x + 2}}} \right)} \right] + c \hfill \\

      = \frac{1}<br />
{2}\tan ^{ - 1} \left( {x + 1} \right) + \frac{{\left( {x + 1} \right)}}<br />
{{\left( {x^2  + 2x + 2} \right)^2 }} + c \hfill \\ <br />
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  4. #4
    Junior Member symstar's Avatar
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    I was wondering why your answer differed slightly from my books answer and I found the problem here...

    Quote Originally Posted by Shyam View Post
       = \int {\frac{{1 + \cos 2\theta }}<br />
{2}} d\theta  \hfill \\

       = \frac{1}<br />
{2}\left( {\theta  + \sin 2\theta } \right) + c \hfill \\
    However, I believe that if  dv=\cos 2\theta d\theta then  v=\tfrac{1}{2}\sin 2\theta .
    This will make the final answer: <br />
= \frac{1}<br />
{2} \left[\tan ^{ - 1} \left( {x + 1} \right) + \frac{{\left( {x + 1} \right)}}<br />
{{\left( {x^2 + 2x + 2} \right)^2 }}\right] + c \hfill \\<br />

    I also don't understand how you did this step, could you go into more detail please?:

    Quote Originally Posted by Shyam View Post
     {\text{Also, }}\tan \theta  = \frac{{x + 1}}<br />
{1} \Rightarrow \sin \theta  = \frac{{x + 1}}<br />
{{\left( {x + 1} \right)^2  + 1}} = \frac{{x + 1}}<br />
{{x^2  + 2x + 2}} \hfill \\

     {\text{and also, }} \Rightarrow \cos \theta  = \frac{1}<br />
{{\left( {x + 1} \right)^2  + 1}} = \frac{1}<br />
{{x^2  + 2x + 2}} \hfill \\
    Last edited by symstar; September 9th 2008 at 04:31 PM.
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  5. #5
    Super Member 11rdc11's Avatar
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    \int \frac{1 + cos(2\theta)}{2}~d\theta

    \int \frac{d\theta}{2}~+~\int\frac{cos(2\theta) d\theta}{2}

    Now so you can see this more clearly lets make some substitiutions for the second intergal

    u = 2\theta

    d\theta = \frac{du}{2}


    \int\frac{cos(2\theta) d\theta}{2} = \frac{1}{4}\int cos(u)du = \frac{sin(u)}{4}

    Now back substitute for u

    so this is what you should get

    \frac{\theta}{2} +\frac{sin2\theta}{4} + C

    now you have to back substitute for the other substituion used

    Also using the identity

    sin(2\theta) =2sin\theta cos\theta

    Ok to find out what \theta is you have to remember that we substituted

    x+1 = tan\theta

    so to find \theta you use arctan

    arctan(x+1) = \theta

    now to find out what sin\theta~\text{and}~cos\theta are you need to know

    tan\theta = \frac{\text{opposite}}{\text{adjacent}}

    sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}

    cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}

    You can figure these out because you are given

    tan\theta = \frac{x+1}{1}

    so all you have to do is find out what the hypotenuse is since you are already given the opposite and adjacent

    H^2 = (x+1)^2 +1^2

    H = \sqrt{x^2 + 2x +2}

    So after alot of typing lol your final integral should look like this

    \frac{arctan(x+1)}{2} + \frac{x+1}{2x^2+4x+4} + C
    Last edited by 11rdc11; September 9th 2008 at 06:03 PM.
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  6. #6
    Junior Member symstar's Avatar
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    Right, 11rdc11. I was getting fouled up because Shyam had not integrated \int \frac{1 + cos(2\theta)}{2}~d\theta correctly and it took me a bit to catch this.
    Also, the values he put for \sin\theta and \cos\theta should have included sqrt in the denominator.

    In the end, I can only say that it helped me hone some of my math skills

    But, thanks to both of you for the help! I finally figured it out!

    Final answer: \frac{1}{2} \left[ \tan^{-1}{(x+1)} + \frac{x+1}{x^2+2x+2}\right] + C
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