Trig Substitution Question (Calculus II)

So I have to evaluate the indefinite integral of sqrt(4x^2 + 9)/x^4 dx.

A similar example says the substitution to use is x = (3/2)tan(theta), but I really don't understand why that is. I would think it would be x = (2/3)tan(theta), if you factor the 9 out of everything within the square root so that you end up with the tan+1 = sec identity.

∫√(4x^2 + 9) dx / x^4

let 4x^2 = 9 tan^2(t) => x^2 = (9/4)tan^2(t) ==> tan t = (2/3)x
2x = 3 tan(t)
2dx = 3 sec^2(t) dt
dx = (3/2) sec^2(t) dt

∫√(9tan^2(t) + 9) (3/2) sec^2(t) dt / (81/16) tan^4(t)

= ∫3√( 1 + tan^2(t)) (3/2) sec^2(t) dt / (81/16) tan^4(t)

= 9/2 ∫ sec^3(t) dt / (81/16) tan^4(t)

= 8/9∫ sec^3(t) dt / tan^4(t)

= 8/9∫ cos (t) dt / sin^4(t)

= 8/9∫ sin^(-4) t d( sin t )

= - (8/27) (1 / sin^3(t) + C

= - (8/27) (1 / sin^3(t) + C ( recall that sin t = tan(t) / sec(t)

= - (8/27) ( sec^3(t) / tan^3(t) + C

= - (8/27) ( 1 + tan^2(t))^3/2 / tan^3(t) + C

substitute tan(t) = (2/3) x

= - (8/27) ( 1 + 4x^2/9))^3/2 /(8x^3/27) + C

= - (8/27) ( 9 + 4x^2))^3/2 / 27(8x^3/27) + C

= - (1/27) ( 9 + 4x^2))^3/2 / x^3 + C

Recall the pythagorean identity:

sec(t)^2 - tan(t)^2 = 1

It would follow that:

r * sec(t)^2 - r * tan(t)^2 = r
r * sec(t)^2 = r + r * tan(t)^2
r * sec(t)^2 = r * (1 + tan(t)^2)

So,

4x^2 + 9 = 9 * (1 + tan(t)^2)
4x^2 + 9 = 9 + 9tan(t)^2
4x^2 = 9tan(t)^2
2x = 3 * tan(t)
x = (3/2) * tan(t)
dx = (3/2) * sec(t)^2 * dt

So:

sqrt(4x^2 + 9) * dx / x^4 =>
sqrt(9 + 9 * tan(t)^2) * (3/2) * sec(t)^2 * dt / ((3/2) * tan(t))^4 =>
sqrt(9sec(t)^2) * (3/2) * sec(t)^2 * dt / ((3/2)^4 * tan(t)^4) =>
3sec(t) * sec(t)^2 * dt / ((3/2)^3 * tan(t)^4) =>
3 * (1/cos(t)^3) * dt / ((27/8) * (sin(t)^4 / cos(t)^4)) =>
3 * (8/27) * (1/cos(t)^3) * dt * (cos(t)^4 / sin(t)^4)) =>
(8/9) * (cos(t) * dt / sin(t)^4

u = sin(t)
du = cos(t) * dt

(8/9) * du / u^4

Integrate

(8/9) * (-1/3) / u^3 + C =>
(-8/27) / sin(t)^3 + C =>
-8 / (27 * sin(t)^3) + C

x = (3/2) * tan(t)
x = 3 * sin(t) / (2 * cos(t))
2 * cos(t) * x = 3 * sin(t)
4 * cos(t)^2 * x^2 = 9 * sin(t)^2
4 * (1 - sin(t)^2) * x^2 = 9 * sin(t)^2
4x^2 - 4 * sin(t)^2 * x^2 = 9 * sin(t)^2
4x^2 = sin(t)^2 * (9 + 4x^2)
sin(t)^2 = 4x^2 / (9 + 4x^2)
sin(t) = 2x / sqrt(9 + 4x^2)

-8 / (27 * sin(t)^3) + C
-8 / (27 * 8x^3 / (9 + 4x^2)^(3/2)) + C =>
-8 * (9 + 4x^2)^(3/2) / (27 * 8 *x^3) + C =>
-1 * (9 + 4x^2)^(3/2) / 27 + C =>
(-1/27) * (9 + 4x^2) * sqrt(9 + 4x^2) + C