What is the antiderivative of sqrt(1-x^2)?

Question

I am trying to find the area under the curve using the fundamental theorem of calculus. I need to find the formula to do this.

Answer 1

integral sqrt(1-x^2) dx... let x = sinu, dx =cosudu
=integral cosu.cosu du
= intgeral (cos2u+1)/2 du
= (1/2)sin2u +u/2
= (1/2)2sinu cosu + u/2
= xsqrt(1-x^2) + arcsin(x)/2

Answer 2

This is usually done by a trig substitution:
Let x = sin t, dx = cos t dt.
Then we have to find
∫ cos² t dt.
The simplest way to do this is to use the identity
cos² t = (1/2)*(1 + cos 2t).
(Yes, you can also use parts, but that's a much more
complicated route.)
So we get
∫ cos² t dt = t /2 + sin 2t/4 = (1/2)(t + sin t cos t).
Now we must back substitute. So our final answer is
(1/2)[ arcsin x + x√(1-x²)] + C.

Answer 3

Use trig subsitution, x=cos(theta) and work out the details. Maple tells me you should get

> int(sqrt(1-x^2),x);

2 1/2
1/2 x (1 - x ) + 1/2 arcsin(x)

or 1/2*x*sqrt(1-x^2)+1/2*arcsin(x)+C

Answer 4

the anti-derivative is equal to the integral... so

sqrt(1-x^2) = (1-x^2)^(1/2)....so we take the intergral

integral of (1-x^2)^(1/2) = .5*(1-x^2)*-2x + C if it is an indefinite integral...but if you have a graph...you should define where the integral is from and therefore you should have a definite integral...