Integration Evaluate∫(0 to π/2) cosx(1-cosx)½ dx~~~?

Sol
A=∫(0 to π/2)_Cosx√(1－Cosx)dx
√2A=∫(0 to π/2)_Cosx√(2－-2Cosx)dx
Set y=x/2
x=2y
Cosx=Cos(2y)=2cos^2 y－1
√(2－2Cosx)=√(2－2Cosx)=√(2—2+4Sin^ y)=2Siny
dx=2dy
√2A=∫(0 to π/4)_(2Cos^2 y－1)2Siny2dy
－A/(2√2)=∫(0 to π/4)_(2Cos^2y－1)dCosy
=2Cos^3 y/3－Cosy|(0 to π/4)
=[2*(√2/2)^3/3－√2/2]－(2/3－1)
=(2√2/12－√2/2)+1/3
=(√2/6－3√2/6)+1/3
=1/3－√2/3
A=2√2(√2/3－1/3)=4/3－2√2/3

(1-cosx)/2 or (1-cosx)^(1/2)?