Compute the definite integral:
integral_0^1 sqrt(1-x^2) dx
For the integrand sqrt(1-x^2), substitute x = sin(u) and dx = cos(u) du.
Then sqrt(1-x^2) = sqrt(1-sin^2(u)) = sqrt(cos^2(u)).
This substitution is invertible over 0<u<pi/2 with inverse u = sin^(-1)(x).
This gives a new lower bound u = sin^(-1)(0) = 0 and upper bound u = sin^(-1)(1) = pi/2:
= integral_0^(pi/2) cos(u) sqrt(cos^2(u)) du
Simplify cos(u) sqrt(cos^2(u)) assuming 0<u<pi/2:
= integral_0^(pi/2) cos^2(u) du
Write cos^2(u) as 1/2 cos(2 u)+1/2:
= integral_0^(pi/2) (1/2 cos(2 u)+1/2) du
Integrate the sum term by term and factor out constants:
= 1/2 integral_0^(pi/2) cos(2 u) du+1/2 integral_0^(pi/2) 1 du
For the integrand cos(2 u), substitute s = 2 u and ds = 2 du.
This gives a new lower bound s = 2 0 = 0 and upper bound s = (2 pi)/2 = pi:
= 1/4 integral_0^pi cos(s) ds+1/2 integral_0^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of cos(s) is sin(s):
= (sin(s))/4|_0^pi+1/2 integral_0^(pi/2) 1 du
Evaluate the antiderivative at the limits and subtract.
(sin(s))/4|_0^pi = (sin(pi))/4-(sin(0))/4 = 0:
= 1/2 integral_0^(pi/2) 1 du
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
= u/2|_0^(pi/2)
Evaluate the antiderivative at the limits and subtract.
u/2|_0^(pi/2) = pi/(2 2)-0/2 = pi/4:
Answer: | = pi/4
