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!