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Math Help - another trig sub

  1. #1
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    another trig sub

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

    2x=3tan\theta
    2dx=3sec^2\theta
    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 x=\frac{3}{\sqrt{v^{2}-4}}, whereat dx=-\frac{3v}{\left( v^{2}-4 \right)\sqrt{v^{2}-4}}\,dv, and the integral is -\frac{1}{9}\int{v^{2}\,dv}=-\frac{1}{27}v^{3}+k, thus the original integral equals -\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 x=\frac{3}{2}tan(\theta) then dx=\frac{3}{2}sec^2(\theta). Substituting everything in leaves \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 9tan^2(\theta)+9 to 9sec^2(\theta), you get \frac{8}{9}\int \frac{sec^3(\theta)}{tan^4(\theta)}\,d\theta. This simplifies to \frac{8}{9}\int \frac{cos(\theta)}{sin^4(\theta)}\,d\theta. From here, you can do a u-substitution, letting u=sin(\theta) and du=cos(\theta)d\theta. Plugging in gives \frac{8}{9}\int \frac{\,du}{u^4}. Integrating yields -\frac{8}{27u^3} = (-\frac{2}{3u})^3. Plugging back in sin(\theta) for u gives -(\frac{2}{3sin(\theta)})^3. Since x=\frac{3}{2}tan(\theta), \theta=arctan(\frac{2x}{3}). Plugging back in, you get -(\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 \frac{2x}{3}, the sine of that angle is \frac{2x}{\sqrt{4x^2+9}}. Thus sin(arctan(\frac{2x}{3})) = \frac{2x}{\sqrt{4x^2+9}}. Finally, plugging everything back into the original equation makes: -(\frac{2}{3*\frac{2x}{\sqrt{4x^2+9}}})^3. This simplifies to: -(\frac{\sqrt{4x^2+9}}{3x})^3 or -\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 \sqrt{9}=3 out of \sqrt{9tan^2(x)+9}, just looking at the constants, you have \frac{3}{(\frac{3}{2})^4}*\frac{3}{2}. This simplifies to \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 \frac{8}{9}\int\frac{\sqrt{tan^2(\theta)+1}}{tan^4  (\theta)}*sec^2(\theta)\,d\theta. Turning \sqrt{tan^2(\theta)+1} into sec(\theta) and multiplying the terms yields \frac{8}{9}\int \frac{sec^3(\theta)}{tan^4(\theta)}\,d\theta.

    I hope that helped.
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