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How to integrate (1/x*sqrt(4x^2 -1)dx using parts or trig?

  • Thread starter Hyari
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  • #1
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Homework Statement


(1 / x * sqrt(4x^2 - 1))dx


Homework Equations


done by parts/trig?


The Attempt at a Solution


1 = c
2x = b
sqrt(4x^2 - 1) = a

sin(theta) = 2x

sin(theta) / 2 = x
-(1/2)*cos(theta) d(theta) = dx
cos(theta) = -2*sqrt(4x^2 - 1)

I am not sure what to do about the x?

Can I just replace it all now?
 
Last edited:

Answers and Replies

  • #2
D H
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The substitution you chose is not a particularly good one because [itex]4x^2-1[/itex] transforms to [itex]\sin^2\theta-1=-\cos^2\theta[/itex] with the substitution [itex]\sin\theta=2x[/itex], and [itex]-\cos^2\theta[/itex] is non-positive. (How do you take the square root?) You have [itex]cos\theta=-2\sqrt{4x^2-1}[/itex], which is incorrect.

It is usually better to use some trigonometric substitution other than [itex]\sin\theta[/itex] when addressing equations involving [itex]ax^2-1[/itex].
 
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  • #3
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I have a triangle
/_|

I called the hypotenuse 1, the bottom leg sqrt(4x^2-1) and the right leg 2x. a = theta.

sin(a) = 2x, x = sin(a) / 2, dx = cos(a) / 2 da
sqrt(4x^2 - 1) = cos(a)

replacing dx, x, and sqrt(16x^2 - 1).

I get (cos(a)/2)da / sin(a)/2 * cos(a).

I'm left with 1 / sin(a), integrate csc(a) = -ln[ csc(a) + cot(a) ];

csc = 1 / 2x
cot = sqrt(4x^2 -1) / 2x.

-ln [ (1/2x) + (sqrt(4x^2 - 1) / 2x) ]

Is that right now?

I can't verify because if i plug it into "the integrator" I get arctan o_O
 
  • #4
D H
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You are still using [itex]\sqrt{4x^2-1}=\cos a[/itex] with the substitution [itex]\sin a = 2x[/itex]. That is not correct. You are using the wrong substitution.
 
  • #5
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I don't get what I can do =(.

I looked at the table of integrals.

du / u*sqrt(a^2 - u^2)

spits out a - 1 / a * ln| [ a + sqrt(a^2 -u^2) ] / u | ?
 
  • #6
D H
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http://en.wikipedia.org/wiki/Trigonometric_substitution" [Broken]
 
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  • #7
Gib Z
Homework Helper
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You want 2x to be the hypotenuse! Thats the way to get your radical to be one of the sides! If 1 is the hypotenuse, the radical is sqrt(1-something), and thats not what you want!
 
  • #8
Gib Z
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/_| Hypot- 2x Right leg, radical, bot leg, 1, theta at bottom left
[tex]\cos \theta = \frac{1}{2x}[/tex]

[tex] \sec \theta = 2x[/tex]
[tex]x=\frac{\sec \theta}{2}[/tex]
[tex]dx=\frac{\sec \theta \tan \theta}{2} d\theta[/tex]
[tex]\tan \theta=\sqrt{4x^2 -1}[/tex]

You will get this last step after your correct substitutions.
[tex]\int \frac{dx}{x\sqrt{4x^2-1}}= \theta[/tex]
 
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  • #9
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Holy Moly! That Is A Lot Cleaner! But what does theta = ?

(theta) = -arctan(sqrt(4x^2-1))
 
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  • #10
Gib Z
Homework Helper
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Ahh close, but no negative on the arc tan, since tan theta=sqrt(4x^2-1), arctan both sides,

arc tan(tan theta)=arctan(sqrt4x^2-1)
Which simplifies to

theta=arctan(sqrt(4x^2-1))
 

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