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slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1To 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: \[{1 \over 2} \cos (2x  3) + C\]

bmelyk
 2 years ago
Best ResponseYou've already chosen the best response.0i had positive 1/3 not 1/2

slaaibak
 2 years ago
Best ResponseYou've already chosen the best response.1How did you get a 1/3 ?

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

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

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