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Math Help - Am I on the right track?: Trigonometric Substitution

  1. #1
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    Am I on the right track?: Trigonometric Substitution

    Use the indicated substitution to evaluate the integral

    1
    (integral) dx/(x^2)sqrt{(x^2)+4), x=2tan(θ)
    1/2

    first got the derivative of x=2tan(θ)-->dx=2sec^2 (θ) dθ

    then I evaluated the denominator, which becomes 8tan^2(θ)secθ

    1
    (integral) dx/(x^2)sqrt{(x^2)+4) =
    1/2

    integral 1,1/2: 2sec^2(θ)dθ/8tan^2(θ)sec(θ) =

    (1/4) integral 1,1/2: 1dθ/cos(θ)(sin(θ)/cos(θ))(tan(θ))

    (1/4) integral 1,1/2: 1/sin(θ)tan(θ) dθ

    (1/4 integral 1,1/2: csc(θ) dθ) x (1/4 integral 1,1/2: cot(θ) dθ)

    1
    (1/4)(ln|csc(θ)-cot(θ)|) x (ln|sin(θ)|) + C
    1/2

    Thanks!
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  2. #2
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    Hello, wyhwang7!

    You went off near the end . . .


    Use the indicated substitution to evaluate the integral.
    . . . \int^1_{\frac{1}{2}} \frac{dx}{x^2\sqrt{x^2+4}} \quad x = 2\tan\theta

    First got the derivative of: x\,=\,2\tan\theta \quad\Rightarrow\quad dx \,=\,2\sec^2\!\theta\,d\theta

    then I evaluated the denominator, which becomes: 8\tan^2\!\theta\sec\theta

    \int \frac{dx}{x^2\sqrt{x^2+4}} \;= \; \int \frac{2\sec^2\!\theta\,d\theta}{8\tan^2\!\theta\se  c\theta} \;=\;\frac{1}{4}\int\frac{\sec\theta\,d\theta}{\ta  n^2\!\theta}


    We have: . \frac{1}{4}\int\frac{\left(\frac{1}{\cos\theta}\ri  ght)}{\left(\frac{\sin^2\!\theta}{\cos^2\!\theta}\  right)} \,d\theta \;=\;\frac{1}{4}\int\frac{\cos\theta}{\sin^2\!\the  ta}\,d\theta

    . . =\;\frac{1}{4}\int\frac{1}{\sin\theta}\cdot\frac{\  cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{4}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{4}\csc\theta


    Back-substitute: . 2\tan\theta \:=\:x\quad\Rightarrow\quad \tan\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{adj}

    . . Then: . hyp \,=\,\sqrt{x^2+4} \quad\Rightarrow\quad \csc\theta \:=\:\frac{\sqrt{x^2+4}}{x}


    Now we must evaluate: . -\frac{1}{4}\cdot\frac{\sqrt{x^2+4}}{x}\,\bigg]^1_{\frac{1}{2}}

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