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Thread: trig substitution

  1. #1
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    May 2010
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    WI - USA
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    trig substitution

    Good afternoon all,

    I am looking at what I hope is my last trig substitution question:

    $\displaystyle
    \int_{0}^{1}{\frac{1}{x^2+2x+2}dx
    $

    Through completing the square I have:

    $\displaystyle
    \int_{0}^{1}{\frac{1}{(x+1)^2+1}dx
    $

    My chosen substitution is:

    $\displaystyle x = \sqrt{1}tan\theta = tan\theta$
    and
    $\displaystyle
    x = \sqrt{1}sec^2\theta d\theta = sec^2\theta d\theta
    $

    Using this substitution I get:

    $\displaystyle
    (x+1)^2 = (tan\theta + 1)^2 = tan^2\theta + 1 = sec^2\theta
    $

    For the new terminals of integration I get:

    $\displaystyle 1 = tan\theta , \theta = tan^{-1}0 = \frac{\pi}{4}$

    $\displaystyle 0 = tan\theta , \theta = tan^{-1}1 = 0$

    Subbing into the origanl integral:

    $\displaystyle
    \int_{tan^{-1}1}^{tan^{-1}0}{\frac{1}{sec^2\theta+1}sec^2\theta d\theta}
    $

    $\displaystyle
    (1)\int_{tan^{-1}1}^{tan^{-1}0}{\frac{sec^2\theta }{sec^2\theta+1}d\theta}
    $

    $\displaystyle
    \int_{tan^{-1}1}^{tan^{-1}0}{\theta d \theta}
    $

    $\displaystyle \frac{\theta^2}{2}$ from $\displaystyle 0$ to $\displaystyle \frac{\pi}{4}$

    Resulting in:

    $\displaystyle \frac{(\frac{\pi}{4})^2}{2}-\frac{0}{2} = \frac{\pi^2}{32}$

    Checking with my calculator shows that I should have come to:

    $\displaystyle
    \frac{\pi}{4}-tan^{-1}(\frac{1}{2})
    $

    My suspicion is that I mishandled the 1 in the denominator. Am I correct in this thinking?
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  2. #2
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    I think that you have gone too far. Simple substitution x'=x+1 will solve the problem in a few moments (clue: arctan(x)).
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  3. #3
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    I'm not sure I follow you.
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  4. #4
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    let's go over it. x'=x+1. dx'=dx. Your integral becomes integral over (1/(1+x'^2)) when x' is between 1 to 2, which gives you immediately arctan(2)-arctan(1).

    I have just assumed that you are familiar with the fact that $\displaystyle \int_{a}^{b}{\frac{1}{1+x^2}dx=atan(b)-atan(a).$
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  5. #5
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    Lexington, MA (USA)
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    Hello, MechEng!

    Your substitution and subsequent algebra is off . . .


    $\displaystyle \displaystyle{\int_{0}^{1}{\frac{dx}{x^2+2x+2}$

    We have: .$\displaystyle \displaystyle{I \;=\; \int^1_0 \frac{dx}{(x+1)^2+1}}$


    Let $\displaystyle x+1 \:=\:\tan\theta \quad\Rightarrow\quad dx \:=\:\sec^2\!\theta\,d\theta$

    . . and: .$\displaystyle (x+1)^2+1 \:=\:\tan^2\!\theta+1 \:=\:\sec^2\!\theta$


    We have: .$\displaystyle \tan\theta \:=\:x+1$

    . . When $\displaystyle x = 0\!:\;\tan\theta \:=\:1 \quad\Rightarrow\quad \theta \,=\,\frac{\pi}{4}$

    . . When $\displaystyle x = 1\!:\;\tan\theta \:=\:2 \quad\Rightarrow\quad \theta \,=\,\tan^{\text{-}1}\!2$


    Substitute: .$\displaystyle \displaystyle{I \;=\;\int^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}}\frac{\sec^2\!d\theta}{\sec ^2\!\thetta} \;=\;\int^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}}d\theta }$


    . . . . . . . . . . .$\displaystyle =\;\theta\,\bigg]^{\tan^{\text{-}1}\!2}_{\frac{\pi}{4}} \;=\; \tan^{\text{-}1}\!2 - \dfrac{\pi}{4} $
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