# Arcsec question

1. Apr 11, 2010

### lude1

1. The problem statement, all variables and given/known data
Can this integral be evaluated using the basic integration formulas?

1/{x[sqrt(1-(x^2))]}dx
Answer in the back of the book: No

2. Relevant equations
[1/a]arcsecc(|u|/a) + C

3. The attempt at a solution
The reason why it cannot be an arcsec is because for the denominator, it is not the square root of (u^2 - a^2) aka: (x^2) - 1.

Thus, I wanted to take a negative out to make it -(x^2) + 1. Then I would take the negative out of the square root, then place the negative before the integral sign. Then, the answer would be -arcsec.

However, because the answer says I cannot do this problem, that means I did something wrong. However, if I did this example wrong, that means I got two other homework problems wrong. However, compared to the answers in the back of the book, I got the correct answer using this method (pull and negative out).

Therefore, I was wondering - am I doing something wrong, or did the book print a mistake? If I did something wrong, that means I'll need help with my other two homework problems :/

2. Apr 11, 2010

### LCKurtz

The expression sqrt(1-x2) is only defined for x between -1 and 1, assuming you aren't using complex numbers. If you could factor it as you say, your square root would look like sqrt(x2-1) which has domain |x| ≥ 1. Different function, different domain.

3. Apr 11, 2010

### Cyosis

Not sure if this counts as a basic integration formula, but you can use the substitution u=sin(x) to get the correct answer.

4. Apr 11, 2010

### lude1

If that is the case, then I was wondering if you could help me with this integration problem?

1. The problem statement, all variables and given/known data.

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

2. Relevant equations
arcsinx

3. The attempt at a solution

I completed the square on the bottom resulting with

dx/sqrt(-(x+2)^2 - 4)

I originally took out the negative (thus having sqrt(a^2 - u^2)). My answer came out to be the same as the back of the book

arcsin[(x+2)/2] + C

However, if this is the same as the problem above, I can't take out the negative. Then, I'm stuck after completing the square.

5. Apr 11, 2010

### LCKurtz

-x2-4x = -(x2+4x + 4)+4 = -(x+2)2+4

so this gives you a 1 - u2 form under the radical, which should give you an arcsine. You did two things wrong and got lucky.

6. Apr 11, 2010

### Cyosis

Could you show us exactly how you are managing to move the -1 out of a root?

7. Apr 11, 2010

### lude1

Ooh I see.. wow. Thanks a lot!