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[tex]\int \frac{sqrt(9x^2-4)}{x}dx[/tex]
I'm stuck at this one part, but i'll show you my steps so you guys can see if I'm doing it right or not.
factor out the 9...
[tex]\int \frac{sqrt(9(x^2-4/9)}{x}dx[/tex]
x = 2/3*sec(theta)
dx = 2/3*sec(theta)tan(theta) d(theta)
then i just subbed 2/3*sec(theta) for one of the X first.
[tex]sqrt(9(4/9*sec(theta)^2 - 4/9))[/tex] => [tex]sqrt(4*sec(theta)^2-4[/tex]
=> [tex]sqrt(4(sec(theta)^2-1))[/tex]
using trig. id...
=> 2tan(theta)
now i will sub in for the other x and dx.
[tex]\int \frac{2*tan(theta)}{2/3*sec(theta)}*2/3*sec(theta)*tan(theta)[/tex]
2/3*sec(theta) canceles out each other so...
[tex]2\int tan(theta)^2[/tex]
=> trig id... [tex]2\int sec(theta)^2-1[/tex]
well i know the integral of sec(theta)^2 is just tan(theta). but what is the integral of 1? it would be x in other cases, but what would the integral of 1 be in this case?
I'm stuck at this one part, but i'll show you my steps so you guys can see if I'm doing it right or not.
factor out the 9...
[tex]\int \frac{sqrt(9(x^2-4/9)}{x}dx[/tex]
x = 2/3*sec(theta)
dx = 2/3*sec(theta)tan(theta) d(theta)
then i just subbed 2/3*sec(theta) for one of the X first.
[tex]sqrt(9(4/9*sec(theta)^2 - 4/9))[/tex] => [tex]sqrt(4*sec(theta)^2-4[/tex]
=> [tex]sqrt(4(sec(theta)^2-1))[/tex]
using trig. id...
=> 2tan(theta)
now i will sub in for the other x and dx.
[tex]\int \frac{2*tan(theta)}{2/3*sec(theta)}*2/3*sec(theta)*tan(theta)[/tex]
2/3*sec(theta) canceles out each other so...
[tex]2\int tan(theta)^2[/tex]
=> trig id... [tex]2\int sec(theta)^2-1[/tex]
well i know the integral of sec(theta)^2 is just tan(theta). but what is the integral of 1? it would be x in other cases, but what would the integral of 1 be in this case?