Question

Integrate the following:

Answer 1

\[∫ \sin(2x-3)dx\]

Answer 2

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:

\[-{1 \over 2} \cos (2x - 3) + C\]

Answer 3

i had positive 1/3 not -1/2

Answer 4

How did you get a 1/3 ?

Answer 5

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

Answer 6

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

Answer 7

i meant 2-3x lol sorry!

Answer 8

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.

Answer 9

yay : )