Jika integral dari (sin2x-cos2x)dx = 1/2^[1/2] sin(2x-a)+b, maka carilah 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 dan sin a= -1/√2
cosa=sina=-1/√2
a=5pi/4
ambil LHS
1/√2 sin(2x+5pi/4) +c
= 1/√2 sin(2x-a)+b
a=-5pi/4 dan b=konstan.
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