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

1. Feb 24, 2007

### Hyari

1. The problem statement, all variables and given/known data
(1 / x * sqrt(4x^2 - 1))dx

2. Relevant equations
done by parts/trig?

3. 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: Feb 24, 2007
2. Feb 24, 2007

### D H

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

It is usually better to use some trigonometric substitution other than $\sin\theta$ when addressing equations involving $ax^2-1$.

Last edited: Feb 24, 2007
3. Feb 24, 2007

### Hyari

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

4. Feb 24, 2007

### D H

Staff Emeritus
You are still using $\sqrt{4x^2-1}=\cos a$ with the substitution $\sin a = 2x$. That is not correct. You are using the wrong substitution.

5. Feb 24, 2007

### Hyari

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. Feb 24, 2007

### D H

Staff Emeritus
http://en.wikipedia.org/wiki/Trigonometric_substitution" [Broken]

Last edited by a moderator: May 2, 2017
7. Feb 24, 2007

### Gib Z

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. Feb 24, 2007

### Gib Z

/_| Hypot- 2x Right leg, radical, bot leg, 1, theta at bottom left
$$\cos \theta = \frac{1}{2x}$$

$$\sec \theta = 2x$$
$$x=\frac{\sec \theta}{2}$$
$$dx=\frac{\sec \theta \tan \theta}{2} d\theta$$
$$\tan \theta=\sqrt{4x^2 -1}$$

You will get this last step after your correct substitutions.
$$\int \frac{dx}{x\sqrt{4x^2-1}}= \theta$$

Last edited: Feb 24, 2007
9. Feb 24, 2007

### Hyari

Holy Moly! That Is A Lot Cleaner! But what does theta = ?

(theta) = -arctan(sqrt(4x^2-1))

Last edited: Feb 24, 2007
10. Feb 24, 2007

### Gib Z

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))