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Thread: another trig sub

  1. #1
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    another trig sub

    $\displaystyle \int \frac{\sqrt{4x^2+9}}{x^4}dx = \int \frac{\sqrt{(2x)^2+3^2}}{x^4}dx$

    $\displaystyle 2x=3tan\theta$
    $\displaystyle 2dx=3sec^2\theta$
    $\displaystyle dx=\frac{3}{2}sec^2\theta$

    Am I going in the right direction? And what do I put back in for the denominator??
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  2. #2
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    Krizalid's Avatar
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    You really need to use trig. sub.? 'cause I'd put $\displaystyle x=\frac{3}{\sqrt{v^{2}-4}},$ whereat $\displaystyle dx=-\frac{3v}{\left( v^{2}-4 \right)\sqrt{v^{2}-4}}\,dv,$ and the integral is $\displaystyle -\frac{1}{9}\int{v^{2}\,dv}=-\frac{1}{27}v^{3}+k,$ thus the original integral equals $\displaystyle -\frac{1}{27}\cdot \left( 4+\frac{9}{x^{2}} \right)^{3/2}+k.$
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  3. #3
    Super Member redsoxfan325's Avatar
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    If you let $\displaystyle x=\frac{3}{2}tan(\theta)$ then $\displaystyle dx=\frac{3}{2}sec^2(\theta)$. Substituting everything in leaves $\displaystyle \int \frac{\sqrt{9tan^2(\theta)+9}}{(\frac{3}{2}tan(\th eta))^4}*\frac{3}{2}sec^2(\theta)\,d\theta$. After a bit of simplification, including changing $\displaystyle 9tan^2(\theta)+9$ to $\displaystyle 9sec^2(\theta)$, you get $\displaystyle \frac{8}{9}\int \frac{sec^3(\theta)}{tan^4(\theta)}\,d\theta$. This simplifies to $\displaystyle \frac{8}{9}\int \frac{cos(\theta)}{sin^4(\theta)}\,d\theta$. From here, you can do a u-substitution, letting $\displaystyle u=sin(\theta)$ and $\displaystyle du=cos(\theta)d\theta$. Plugging in gives $\displaystyle \frac{8}{9}\int \frac{\,du}{u^4}$. Integrating yields $\displaystyle -\frac{8}{27u^3} = (-\frac{2}{3u})^3$. Plugging back in $\displaystyle sin(\theta)$ for $\displaystyle u$ gives $\displaystyle -(\frac{2}{3sin(\theta)})^3$. Since $\displaystyle x=\frac{3}{2}tan(\theta)$, $\displaystyle \theta=arctan(\frac{2x}{3})$. Plugging back in, you get $\displaystyle -(\frac{2}{3sin(arctan(\frac{2x}{3}))})^3$. If you draw out the triangle, you will see that if the tangent of an angle is $\displaystyle \frac{2x}{3}$, the sine of that angle is $\displaystyle \frac{2x}{\sqrt{4x^2+9}}$. Thus $\displaystyle sin(arctan(\frac{2x}{3})) = \frac{2x}{\sqrt{4x^2+9}}$. Finally, plugging everything back into the original equation makes: $\displaystyle -(\frac{2}{3*\frac{2x}{\sqrt{4x^2+9}}})^3$. This simplifies to: $\displaystyle -(\frac{\sqrt{4x^2+9}}{3x})^3$ or $\displaystyle -\frac{(4x^2+9)^{\frac{3}{2}}}{27x^3} + C$.


    That problem was a huge pain to do with a trig sub, but it is right - I checked it on Maple.
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  4. #4
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    how did u simplify the bottom of the fraction (3/2)tan(theta)^4 to tan^4(theta). Sorry, I don't understand.
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  5. #5
    Super Member redsoxfan325's Avatar
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    I combined all the constants. Since you can pull a $\displaystyle \sqrt{9}=3$ out of $\displaystyle \sqrt{9tan^2(x)+9}$, just looking at the constants, you have $\displaystyle \frac{3}{(\frac{3}{2})^4}*\frac{3}{2}$. This simplifies to $\displaystyle \frac{9}{\frac{81}{16}*2} = \frac{9}{\frac{81}{8}} = \frac{1}{\frac{9}{8}} = \frac{8}{9}$, which I pulled outside of the integral as a constant, leaving the integral $\displaystyle \frac{8}{9}\int\frac{\sqrt{tan^2(\theta)+1}}{tan^4 (\theta)}*sec^2(\theta)\,d\theta$. Turning $\displaystyle \sqrt{tan^2(\theta)+1}$ into $\displaystyle sec(\theta)$ and multiplying the terms yields $\displaystyle \frac{8}{9}\int \frac{sec^3(\theta)}{tan^4(\theta)}\,d\theta$.

    I hope that helped.
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