Remember this result:
cos2x = 1 - 2sinxsinx
sinxsinx = 1/2(1-cos2x)
integral sign = }
i)
A = } pi/2 , 0 , sinxsinx . dx
= } pi/2 , 0 , 1/2(1-cos2x) . dx
= 1/2 } pi/2 , 0 , (1-cos2x) . dx
= 1/2 [x - 1/2sin2x] , pi/2 , 0
= 1/2 [(pi/2 - 1/2sinpi) - (0 - 1/2sin0)]
= 1/2 [pi/2...