# Trig sub. integrals

$$\int \frac{sqrt(9x^2-4)}{x}dx$$

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...
$$\int \frac{sqrt(9(x^2-4/9)}{x}dx$$

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.

$$sqrt(9(4/9*sec(theta)^2 - 4/9))$$ => $$sqrt(4*sec(theta)^2-4$$

=> $$sqrt(4(sec(theta)^2-1))$$

using trig. id.....

=> 2tan(theta)

now i will sub in for the other x and dx.

$$\int \frac{2*tan(theta)}{2/3*sec(theta)}*2/3*sec(theta)*tan(theta)$$

2/3*sec(theta) canceles out each other so...

$$2\int tan(theta)^2$$

=> trig id... $$2\int sec(theta)^2-1$$

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?

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vsage
Assuming you did the rest of the work right, you're integrating with respect to theta so $$\int {1 d\theta}$$ is $$\theta$$

yea i knew that, but i didnt know how to sub in for theta. i can sub in for tan(theta) by drawing a right triangle.

tan(theta) = $$\frac{sqrt(9x^2-4)}{2}$$

so the answer should be $$2*\frac{9x^2-4}{2} - 2(theta)$$

what should i sub in for theta?

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$$\int [a(x)+b(x)]dx = \int a(x) dx +\int b(x) dx$$

Intergration of 1 will be theta (not x)

$$\theta = cos^{-1}\frac{2}{3x}$$

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dextercioby
Homework Helper
$$\frac{3x}{2}=\cosh u$$ ?

I think it comes out nicely...

Daniel.

Gamma said:
$$\int [a(x)+b(x)]dx = \int a(x) dx +\int b(x) dx$$

Intergration of 1 will be theta (not x)

$$\theta = cos^{-1}\frac{2}{3x}$$
$$sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}$$ (which looks extactly like my answer if i subbed in $$arctan\frac{2}{sqrt(9*x^2-4)}$$ for theta)
i used a computer to find the answer, but i would like to know how to find theta. can someone help?

how did you get $$\theta = cos^{-1}\frac{2}{3x}$$

i tried to sub in what you got for theta, but it told me that i was wrong (i submit my hw online and it checks my stuff)

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if you can draw the triangle, write something like $$\cos \theta = (something)$$ and take the inverse cosine of both sides.

learningphysics
Homework Helper
x = 2/3*sec(theta)
Just solve for theta from the above equation.

x=2/3*sec(theta)
=>
3/2x=sec(theta)
1/sec*3/2x=theta
sec^(-1)*3/2x = theta

i got that, am i doing something wrong?

learningphysics
Homework Helper
yea i knew that, but i didnt know how to sub in for theta. i can sub in for tan(theta) by drawing a right triangle.

tan(theta) = $$\frac{sqrt(9x^2-4)}{2}$$

so the answer should be $$2*\frac{9x^2-4}{2} - 2(theta)$$

what should i sub in for theta?
$$\sqrt{9x^2-4} - 2(arccos(2/3x))$$
?

learningphysics
Homework Helper
x=2/3*sec(theta)
=>
3/2x=sec(theta)
It should be 3x/2 = sec(theta)

the correct answer is $$sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}$$

that means that if theta = $$arctan\frac{2}{sqrt(9*x^2-4)}$$

i would get the correct answer

on the other hand, if i subbed in theta = $$\theta = cos^{-1}\frac{2}{3x}$$

i would get the wrong answer

learningphysics
Homework Helper
It seems like the way you solved it, the answer comes out to:

$$sqrt((9*x^2-4))-2*arctan\frac{sqrt(9*x^2-4)}{2}$$

$$sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}$$

But I can't find the source of the discrepancy. Can somebody see the reason? I can't find any errors in your work. All I can think of is tan(theta)=1/tan(Pi/2-theta). Maybe somehow this caused a problem somewhere?? :yuck:

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dextercioby
$$\frac{2}{3x}=\sin t$$ ?