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

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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!
 

Answers and Replies

  • #2
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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 [itex]f^\prime(x)=2x[/itex] with f(0)=1. Then [itex]f(x)=x^2+c[/itex] and [itex]x^2+1+c[/itex] 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 [itex]f(x)=x^2+c_1[/itex] and [itex]x^2+1+c_2[/itex].

And in your case, it's better to say f(x)=4arcsin(x)+c1 and f(x)=-4arcos(x)+c2.
 
  • #3
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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 [itex]f^\prime(x)=2x[/itex] with f(0)=1. Then [itex]f(x)=x^2+c[/itex] and [itex]x^2+1+c[/itex] 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 [itex]f(x)=x^2+c_1[/itex] and [itex]x^2+1+c_2[/itex].

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!
 

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