Question:

What's the Primitive function of:

f'(x) = -1/((1-x^2)^0.5)

... If f(x) = -arcsin(x), then f'(x) = -1/sqrt(1 - x^2)

... f(x) = -arcsin(x) + c for any constant c.

Anyone here a bit smarter than me?

I'm not going to attempt answering the problem (in part because
I haven't seen this yet) but I just want to point something out, in
case you've missed it:

(1-x^2)^0.5 is the same as sqrt(1 - x^2)

Hope that helps ? ;)

If not, don't mind me... :p

Do you have a "Schaum's Outline"? It'll probably be in there. If nobody comes up with it by tonight I'll dig mine out and find it.

Didn't you.... answer your own question?:bugeye:

Man, that's the question.. was I right or did I just sign my own death..?

YES. What everyone is saying is:

Since (1-x^2)^(1/2) is the same as [sqrt](1-x^2),

The anti-derivative of both is - arcsin(x)+ C.

The antiderivative of 1/sqrt(1-x²) is arcsin x.

Isn't the antiderivative of -1/sqrt(1-x²) equal to arccos x?

Yeah... it is... :o

Thanks:) Then I might be around a bit longer..

Originally posted by Icebird
Question:

What's the Primitive function of:

f'(x) = -1/((1-x^2)^0.5)

... If f(x) = -arcsin(x), then f'(x) = -1/sqrt(1 - x^2)

... f(x) = -arcsin(x) + c for any constant c.

Anyone here a bit smarter than me?



Do you know how to actually evaluate the integral?
See attached.

wait...
Ii I sum up all the above posts...

arccos(x)=-arcsin(x)    ?

...how come?

Originally posted by BloodSuckingGerbile
wait...
Ii I sum up all the above posts...

arccos(x)=-arcsin(x)    ?

...how come?

you re forgetting the constant of integration. arccos x = pi/2 - arcsin x

Oh, right. Thanks. Got it.