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Integrate the following:

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\[∫ \sin(2x-3)dx\]
To make it a bit easier, let's use substitution: \[u = 2x - 3 \rightarrow du = 2 dx \rightarrow dx = {1 \over 2} du\] Now the integral becomes: \[\int\limits {1 \over 2} \sin {u} du = -{1 \over 2}\cos u + C\] substituting u=2x-3 back: \[-{1 \over 2} \cos (2x - 3) + C\]
i had positive 1/3 not -1/2

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Other answers:

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!
  • hba
Well if that is the case your answer would be \[\frac{ 1 }{ 3 }\cos(2-3x)+C\] Which you said you got. Therefore,you are correct.
yay : )

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