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\[∫ \sin(2x-3)dx\]

i had positive 1/3 not -1/2

How did you get a 1/3 ?

\[-1/3∫ \sin(2-3x)d(2-3x) ... 1/3\cos(2-3x)+c\]

but in the problem you stated it's 2x - 3, and not 2 - 3x?

i meant 2-3x lol sorry!

yay : )