Verified answer

Well, first note that 1 - sin^2(x) = cos^2(x).

Then,√(cos^2(x)) = |cos(x)|.

∫|cos(x)| dx

on the given interval can be rewritten as the sum of

∫cos(x) dx on the interval [0, π/2]

and

∫-cos(x) dx on the interval [π/2, π]

Solving for this, we get 2.

Of (sin²(x) + cos²(x) = 1) :

cos²(x) = 1- sin²(x)

∫√(1-sin²(x) ).dx =

∫√cos²(x).dx =

∫l cos(x) l.dx =

l sin(x) l + C =

-sin(x) or sin(x)

the intervel is [0,π], consequently

∫√(1-sin²(x) ).dx = sin(x) + C

sqrt(1-sin^2x)=cosx

int(cosxdx)=sinx(0,pi)=

=0