How do you evaluate #int sqrt (1-x^2) dx # for [-1, 1]?

1 Answer
Sep 19, 2015

#pi/2#

Explanation:

let #x=sintheta#
#sqrt(1-x^2)=sqrt(1-sin^2theta)=abs(costheta)#
in #-pi/2 to pi/2# #cos theta# is positive so #abscostheta =costheta#
changing limits
#if x=-1 then theta =-pi/2#
#if x=1 then theta = pi/2#
#dx=costheta d(theta)#
then
#int_-1^1sqrt(1-x^2)dx# changes to #int_(-pi/2)^(pi/2)cos^2thetad(theta)#
=#int_(-pi/2)^(pi/2)(1+cos2theta)/2d(theta)#
=#int_(-pi/2)^(pi/2)d(theta)/2+int_(-pi/2)^(pi/2)(cos2theta)/2d(theta)#
=#1/2(pi/2-(-pi/2)+1/2(1/2(sin(pi)-sin(-pi)))#
=#pi/2#