# Thread: another trig sub

1. ## 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??

2. 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 $-\frac{1}{27}\cdot \left( 4+\frac{9}{x^{2}} \right)^{3/2}+k.$

3. 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.

4. how did u simplify the bottom of the fraction (3/2)tan(theta)^4 to tan^4(theta). Sorry, I don't understand.

5. 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.