If the integral of (sin2x-cos2x)dx = 1/2^[1/2] sin(2x-a)+b, then find a & b.

∫(sin2x-cos2x)dx = (1/√2)sin(2x-a) + b

∫ sin2xdx – ∫cos2xdx = (1/√2)sin(2x-a) + b

– cos^2x/2-sin^2x/2 + C

= sin2xcosa/√2 – cos2xsina/√2 +b

cos a=-1/√2 and sin a= -1/√2

cosa=sina=-1/√2

a=5pi/4

take LHS

1/√2 sin(2x+5pi/4) +c

= 1/√2 sin(2x-a)+b

a=-5pi/4 and b=constant.

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