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Thread: Trigonometric Substitutions (fixed)

  1. #1
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    Trigonometric Substitutions (fixed)

    (a) If we use the substitution t = 6tan(θ), then which of the following integrals is equivalent to ?

    1















    (b) Use the integral in part (a) to evaluate


    Please show all steps and tank you!
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  2. #2
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    Sorry.
    Can't believe I did that.
    Last edited by chabmgph; Feb 2nd 2009 at 05:17 PM.
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  3. #3
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    Hello, choshiwara!

    (a) If we use the substitution $\displaystyle t = 6\tan\theta$,
    then which of the following integrals is equivalent to: .$\displaystyle \int\frac{dt}{t^2\sqrt{t^2+36}}$

    $\displaystyle (1)\;\int\frac{\sec\theta\,d\theta}{36\tan^2\!\the ta} \qquad (2)\;\int\frac{d\theta}{216\tan\theta\sec\theta} \qquad(3)\;\int\frac{d\theta}{36\tan^2\!\theta} \qquad(4)\;\int\frac{\sec^2\!\theta\,d\theta}{36\t an^2\!\theta} $

    Let: $\displaystyle t \:=\:6\tan\theta \quad\Rightarrow\quad dt \:=\:6\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{t^2+36} \:=\:6\sec\theta$


    Substitute: .$\displaystyle \int\frac{\overbrace{6\sec^2\!\theta\,d\theta}^{dt }}{\underbrace{36\tan^2\!\theta}_{t^2}\cdot\underb race{6\sec\theta}_{\sqrt{t^2+36}}} \;\;=\;\; \int\frac{\sec\theta\,d\theta}{36\tan^2\!\theta}$ . . . answer choice (1)


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    We have: .$\displaystyle \frac{1}{36}\int\frac{\sec\theta}{\tan^2\!\theta}\ ,d\theta \;=\;\frac{1}{36}\int\frac{\frac{1}{\cos\theta}}{\ frac{\sin^2\!\theta}{\cos^2\!\theta}}d\theta \;=\;\frac{1}{36}\int\frac{\cos\theta}{\sin^2\!\th eta}\,d\theta$

    . . $\displaystyle = \;\frac{1}{36}\int\frac{1}{\sin\theta}\!\cdot\frac {\cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{36}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{36}\csc\theta + C$ .[1]


    Back-substitute: .$\displaystyle t \:= \:6\tan\theta \quad\Rightarrow\quad \tan\theta \:=\:\frac{t}{6} \:=\:\frac{opp}{adj}$

    So $\displaystyle \theta$ is in a right triangle with: .$\displaystyle opp = t,\;adj = 6$
    Using Pythagorus, we find that: .$\displaystyle hyp = \sqrt{t^2+36}$
    . . Hence: .$\displaystyle \csc\theta \:=\:\frac{\sqrt{t^2+36}}{t}$


    Therefore, [1] becomes: .$\displaystyle -\frac{\sqrt{t^2+36}}{36t} + C$

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