Trig sub. integrals

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  • #1
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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?
 

Answers and Replies

  • #2
vsage
Assuming you did the rest of the work right, you're integrating with respect to theta so [tex] \int {1 d\theta}[/tex] is [tex]\theta[/tex]
 
  • #3
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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) = [tex]\frac{sqrt(9x^2-4)}{2}[/tex]


so the answer should be [tex]2*\frac{9x^2-4}{2} - 2(theta)[/tex]

what should i sub in for theta?
 
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  • #4
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[tex]\int [a(x)+b(x)]dx = \int a(x) dx +\int b(x) dx[/tex]


Intergration of 1 will be theta (not x)


[tex] \theta = cos^{-1}\frac{2}{3x} [/tex]
 
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  • #5
dextercioby
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How about
[tex]\frac{3x}{2}=\cosh u [/tex] ?

I think it comes out nicely...

Daniel.
 
  • #6
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Gamma said:
[tex]\int [a(x)+b(x)]dx = \int a(x) dx +\int b(x) dx[/tex]


Intergration of 1 will be theta (not x)


[tex] \theta = cos^{-1}\frac{2}{3x} [/tex]
the answer is:
[tex]sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}[/tex] (which looks extactly like my answer if i subbed in [tex]arctan\frac{2}{sqrt(9*x^2-4)}[/tex] 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 [tex] \theta = cos^{-1}\frac{2}{3x} [/tex]

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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  • #7
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if you can draw the triangle, write something like [tex] \cos \theta = (something) [/tex] and take the inverse cosine of both sides.
 
  • #8
learningphysics
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ProBasket said:
x = 2/3*sec(theta)
Just solve for theta from the above equation.
 
  • #9
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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?
 
  • #10
learningphysics
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ProBasket said:
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) = [tex]\frac{sqrt(9x^2-4)}{2}[/tex]


so the answer should be [tex]2*\frac{9x^2-4}{2} - 2(theta)[/tex]

what should i sub in for theta?
So was your final answer:
[tex]\sqrt{9x^2-4} - 2(arccos(2/3x))[/tex]
?
 
  • #11
learningphysics
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ProBasket said:
x=2/3*sec(theta)
=>
3/2x=sec(theta)
It should be 3x/2 = sec(theta)
 
  • #12
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the correct answer is [tex]sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}[/tex]


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

i would get the correct answer


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

i would get the wrong answer
 
  • #13
learningphysics
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It seems like the way you solved it, the answer comes out to:

[tex]sqrt((9*x^2-4))-2*arctan\frac{sqrt(9*x^2-4)}{2}[/tex]

but the correct answer is:

[tex]sqrt((9*x^2-4))+2*arctan\frac{2}{sqrt(9*x^2-4)}[/tex]

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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  • #14
dextercioby
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How about another substitution
[tex] \frac{2}{3x}=\sin t [/tex] ?

Daniel.
 

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