Anti derivative of f'(x)=4/(1-x^2)^(1/2) yields TWO possible answers?

[skyturnred]

Homework Statement

Find the anti derivative of f'(x)=4/(1-x^2)^(1/2) when f(1/2)=1

Homework Equations

The Attempt at a Solution

My problem is as follows: aren't f(x)=4arcsin(x)+c and f(x)=-4arcos(x)+c both perfectly good anti derivatives of f'(x)? In this case, if I plug f(x) in as 1 and x in as (1/2), in the first case I find c to be 1-(4pi)/6, and in the second case I get c to be 1+(4pi)/3. I MUST be wrong somewhere!

Thanks so much for all of your help, so far you guys at physicsforums have been beyond helpful!

 

[micromass]

Homework Statement

Find the anti derivative of f'(x)=4/(1-x^2)^(1/2) when f(1/2)=1

Homework Equations

The Attempt at a Solution

My problem is as follows: aren't f(x)=4arcsin(x)+c and f(x)=-4arcos(x)+c both perfectly good anti derivatives of f'(x)? In this case, if I plug f(x) in as 1 and x in as (1/2), in the first case I find c to be 1-(4pi)/6, and in the second case I get c to be 1+(4pi)/3. I MUST be wrong somewhere!

Notice that (depending on how you define things) arccos(x)=-arcsin(x)+1. So things are equal up to a constant.

Let me do the same thing for another integral, so you can see what is going on. Let f^\prime(x)=2x with f(0)=1. Then f(x)=x^2+c and x^2+1+c are both perfectly good anti derivatives of f'(x). However, if you plug in 0 in the first, then you find c=0. And in the second, you find c=-1.

The solution of course is that you are using two different c's. So you gave two different things the same name. It would be better in my example to say f(x)=x^2+c_1 and x^2+1+c_2.

And in your case, it's better to say f(x)=4arcsin(x)+c₁ and f(x)=-4arcos(x)+c₂.

 

[skyturnred]

Notice that (depending on how you define things) arccos(x)=-arcsin(x)+1. So things are equal up to a constant.

Let me do the same thing for another integral, so you can see what is going on. Let f^\prime(x)=2x with f(0)=1. Then f(x)=x^2+c and x^2+1+c are both perfectly good anti derivatives of f'(x). However, if you plug in 0 in the first, then you find c=0. And in the second, you find c=-1.

The solution of course is that you are using two different c's. So you gave two different things the same name. It would be better in my example to say f(x)=x^2+c_1 and x^2+1+c_2.

And in your case, it's better to say f(x)=4arcsin(x)+c₁ and f(x)=-4arcos(x)+c₂.

Thank you very much that makes sense now!