Evaluating the Arcsec Integral: Can it be Done?

[lude1]

Homework Statement

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

Homework Equations

[1/a]arcsecc(|u|/a) + C

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 :/

 

[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.

 

[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.

 

[lude1]

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

1. Homework Statement .

dx/sqrt(-x^2 - 4x)
​

Homework Equations

arcsinx

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.

 

[LCKurtz]

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

1. Homework Statement .

dx/sqrt(-x^2 - 4x)
​

Homework Equations

arcsinx

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.

-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.

 

[Cyosis]

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

 

[lude1]

Ooh I see.. wow. Thanks a lot!