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bmelyk Group TitleBest ResponseYou've already chosen the best response.0
\[∫ \sin(2x3)dx\]
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
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=2x3 back: \[{1 \over 2} \cos (2x  3) + C\]
 one year ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
i had positive 1/3 not 1/2
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
How did you get a 1/3 ?
 one year ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
\[1/3∫ \sin(23x)d(23x) ... 1/3\cos(23x)+c\]
 one year ago

slaaibak Group TitleBest ResponseYou've already chosen the best response.1
but in the problem you stated it's 2x  3, and not 2  3x?
 one year ago

bmelyk Group TitleBest ResponseYou've already chosen the best response.0
i meant 23x lol sorry!
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Well if that is the case your answer would be \[\frac{ 1 }{ 3 }\cos(23x)+C\] Which you said you got. Therefore,you are correct.
 one year ago
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