I'll suggest to replace the variable x by the function sin
t.
x = sin t
We'll
differentiate both sides and we'll get:
dx = cos t
dt
We'll change the limits of integration,
too:
x = 0 => t = 0
x =
1 => t = pi/2
The definite integral will be
evaluated using the Leibniz Newton formula:
Int sqrt[1 -
(sin t)^2] cos t dt = F(pi/2) - F(0)
But, from Pythagorean
identity, we'll have:
1 - (sin t)^2 = (cos
t)^2
We'll take square root both
sides:
sqrt [1 - (sin t)^2] = sqrt (cos
t)^2
sqrt [1 - (sin t)^2] = cos
t
Int sqrt[1 - (sin t)^2] cos t dt = Int (cos t)^2
dt
Int (cos t)^2 = Int (1+cos
2t)dt/2
Int (1+cos 2t)dt/2 = Int dt/2 + (1/2)*Int cos 2t
dt
Int (1+cos 2t)dt/2 = t/2 + sin
2t/4
F(pi/2) - F(0) = pi/4 + sin pi/4 - 0 + sin
0/4
sin pi = 0
F(pi/2) - F(0)
= pi/4
The definite integral of the given
function is: Int sqrt(1 - x^2) dx = pi/4
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