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Math Help - integration by substitution

  1. #1
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    integration by substitution

    \int\frac{1}{\sqrt{9+4x^2}}dx (Let x=\frac{3}{2}tan\theta)

    \frac{dx}{d\theta}=\frac{3}{2}sec^2\theta

    \int\frac{1}{\sqrt{9+4x^2}}dx=\int\frac{1}{\sqrt{9  +4(\frac{3}{2}tan^2\theta)}} d\theta

    =\frac{3}{18}\int\frac{1+tan^2\theta}{1+tan^2\thet  a}d\theta

    =\frac{3}{18}\theta+c
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  2. #2
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    Re: integration by substitution

    Your substitution is ok. Now calculate:
    \sqrt{9+4x^2}=\sqrt{9+4\left[\frac{3}{2}\tan(\theta)\right]^2}
    =\sqrt{9+9\tan^2(\theta)}=3\cdot \sqrt{1+\tan^2(\theta)}=\frac{3}{\cos(\theta)}

    The integral becomes:
    \int \frac{\frac{3}{2\cos^2(\theta)}}{\frac{3}{\cos(\th  eta)}}d(\theta)
    =\int \frac{\cos(\theta)}{2\cos^2(\theta)}d(\theta)}= \frac{1}{2} \int \frac{d(\theta)}{\cos(\theta)}

    Now you can use the t-formulas or the substitution, let \theta=\arcsin(t).
    Last edited by Siron; July 24th 2011 at 02:35 AM.
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  3. #3
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    Re: integration by substitution

    it will be easier to use hyperbolic substitution
    let x = (3/2) sinh (theta)
    gives dx = (3/2)cosh (theta) d(theta)
    Last edited by waqarhaider; July 24th 2011 at 04:31 AM. Reason: may be I use Latex wrong
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  4. #4
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    Re: integration by substitution

    Quote Originally Posted by Punch View Post
    \int\frac{1}{\sqrt{9+4x^2}}dx (Let x=\frac{3}{2}tan\theta)

    \frac{dx}{d\theta}=\frac{3}{2}sec^2\theta

    \int\frac{1}{\sqrt{9+4x^2}}dx=\int\frac{1}{\sqrt{9  +4(\frac{3}{2}tan^2\theta)}} d\theta

    you forgot to square 3/2 and substitute for dx
    \int\frac{1}{\sqrt{9+4x^2}}dx =

    \int\frac{\frac{3}{2}\sec^2{\theta}}{\sqrt{9+4 \left(\frac{9}{4}\tan^2{\theta} \right)}} \, d\theta =

    \int \frac{3}{2} \cdot \frac{\sec^2{\theta}}{3\sqrt{1 + \tan^2{\theta}}} \, d\theta =

    \frac{1}{2} \int \sec{\theta} \, d\theta =

    \frac{1}{2} \ln|\sec{\theta}+\tan{\theta}| + C

    since x = \frac{3}{2} \tan{\theta} ...

    \tan{\theta} = \frac{2x}{3}

    \sec{\theta} = \sqrt{1 + \frac{4x^2}{9}}

    back substitute ...

    \frac{1}{2} \ln \left|\sqrt{1 + \frac{4x^2}{9}} + \frac{2x}{3} \right| + C
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  5. #5
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    Re: integration by substitution

    Hello, Punch!

    \int\frac{dx}{\sqrt{9+4x^2}}

    \text{Let }2x \,=\, 3\tan\theta \quad\Rightarrow\quad dx \,=\,\tfrac{3}{2}\sec^2\!\theta\,d\theta

    . . \sqrt{9+4x^2} \:=\:\sqrt{9 + 9\tan^2\theta} \:=\;\sqrt{9(1 + \tan^2\!\theta)} \:=\:\sqrt{9\sec^2\!\theta} \:=\:3\sec\theta


    Substitute: . \int \frac{\frac{3}{2}\sec^2\!\theta\,d\theta}{3\sec \theta} \;=\; \tfrac{1}{2}\int\sec\theta\,d\theta \;=\;\tfrac{1}{2}\ln|\sec\theta + \tan\theta| + C


    Back-substitute: . \tan\theta \,=\,\frac{2x}{3},\;\;\sec\theta \,=\,\frac{\sqrt{9+4x^2}}{3}

    We have: . \frac{1}{2}\ln\left|\frac{\sqrt{9+4x^2}}{3} + \frac{2x}{3}\right| + C \;=\;\boxed{\frac{1}{2}\ln\left|\frac{2x + \sqrt{9+4x^2}}{3}\right| + C}


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


    This can be simplified even further . . .

    \tfrac{1}{2}\ln\left|\frac{2x + \sqrt{9+4x^2}}{3}\right| + C

    . . =\;\tfrac{1}{2}\bigg[\ln\left|2x+\sqrt{9+4x^2}\right| - \ln 3\bigg] + C

    . . =\;\tfrac{1}{2}\ln\left|2x + \sqrt{9+4x^2}\right|\, \underbrace{-\, \tfrac{1}{2}\ln 3 + C}_{\text{a constant!}}

    . . =\;\tfrac{1}{2}\ln\left|2x + \sqrt{9+4x^2}\right| + C

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