Integrate Trig Sub. Integrals: $\int \frac{\sqrt{9x^2-4}}{x}dx$

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\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?

 

[vsage]

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

 

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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?

 

[Gamma]

\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}

 

[dextercioby]

How about
\frac{3x}{2}=\cosh u ?

I think it comes out nicely...

Daniel.

 

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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}

the answer is:
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 homework online and it checks my stuff)

 

[kanato]

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

 

[learningphysics]

x = 2/3*sec(theta)

Just solve for theta from the above equation.

 

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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]

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?

So was your final answer:
\sqrt{9x^2-4} - 2(arccos(2/3x))
?

 

[learningphysics]

x=2/3*sec(theta)
=>
3/2x=sec(theta)

It should be 3x/2 = sec(theta)

 

[ProBasket]

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]

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}

but the correct answer is:

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??

 

[dextercioby]

How about another substitution
\frac{2}{3x}=\sin t ?

Daniel.