How do you solve the integral of (1-cos(3t))sin(3t)dt when the lower limit is 0 and the upper is pi/6

Int [(1-cos(3t))sin(3t)]dt=int[sin(3t)-cos(3t)sin(3t)]dt = (-1/3) cos(3t) -(1/2) int[sin(6t)]dt

=(-1/3) cos(3t) +(1/12) cos(6t)

Now calculate in limits

int from 0 to pi/6=(-1/3) cos(pi/2) +(1/12) cos(pi) -[(-1/3)+(1/12)]=0+(1/12)(-1) +1/3 - (1/12)= 1/3 - 2/12 =1/3 -1/6

= 1/6

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