Question

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

Answer 1

`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`