Trig Substitution problem.

[bjackson14]

the problem is (integral of (sqr root(4x^2+9))/x^4) dx
here is what i've done so far:
I simplified the problem first by writing the radical inside the square root as 2 times sqr root(x^2+9/4) because i factored out a 4 which became a 2 on the outside of the square root. next I picked x = (3/2)tan(theta) because in the formula sqr root(x^2+a^2) it simplifies to atan(theta). next i found that dx= (3/2)sec^2(theta)d(theta) because that is the derivative of (3/2)tan(theta). so now i have to solve for sqr root(a^2+x^2) which turns into sqr root(9/4 +(9/4)tan^2(theta)) which simplifies to sqr root(9/4(1+tan^2(theta)) which simplifies again using a trig identity to sqr root (9/4(sec^2(theta)) which now you can solve the sqr root by simplifying it to (3/2)sec(theta).

so now we can rewrite the integral to look like the following:

integral of ((2 * (3/2)sec(theta) * (3/2)sec^2(theta) d(theta))/((3/2)tan(theta))^4 which simplifies to 8/9 * integral of ((sec^3(theta)/tan^4(theta))d(theta) which using more trig identities and writing in terms of sin and cos, i can simplify it to 8/9 * integral of (cos(theta) d(theta)/ sin^4 (theta) d(theta)).

after this im lost because ive tried writing it to look like cot(theta)csc^3(theta) as well as cos(theta)/sin^4(theta). I have tried using u substitution on both all 4 terms and nothing seems to come out right. im pretty confident that im right up to this point but after that im lost.

 

[galactus]

Here is one way. There are always many ways to go about these.

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

Let \(\displaystyle x=\frac{3}{2}tan(t), \;\ dx=\frac{3}{2}sec^{2}(t)dt\)

Making these subs, it simplifies down to

Spoiler: :3gipx1k4

\(\displaystyle \frac{\sqrt{9sec^{2}(t)}}{(\frac{3}{2}tan(t))^{4}}\cdot\frac{3}{2}sec^{2}(t)dt\)

\(\displaystyle =\frac{8}{9}\int\frac{cos(t)}{sin^{4}(t)}dt\)

Now, let \(\displaystyle u=sin(t), \;\ du=cos(t)dt\)

\(\displaystyle \frac{8}{9}\int\frac{1}{u^{4}}du=\frac{-8}{27u^{3}}\)

Resub:

\(\displaystyle \frac{-8}{27}csc^{3}(t)\)

Now, resub again by using \(\displaystyle t=tan^{-1}(\frac{2x}{3})\)

and noting that \(\displaystyle csc(tan^{-1}(x))=\frac{\sqrt{x^{2}+1}}{x}\)

This leads to:

\(\displaystyle \frac{-8}{27}csc^{3}(tan^{-1}(\frac{2x}{3}))\)

\(\displaystyle =\frac{-(4x^{2}+9)^{\frac{3}{2}}}{27x^{3}}\)[/spoiler:3gipx1k4]