∫[-π,π] √[1-(cosx)^2]dx
=∫[-π,0] √[1-(cosx)^2]dx+∫[0,π] √[1-(cosx)^2]dx
=∫[-π,0] [ -(sinx)]dx +∫[0,π] sinxdx
=cos0-cos(-π)+[-cosπ+cos0]
=2+2
=4