## anonymous 3 years ago Integrate the following:

1. anonymous

$∫ \sin(2x-3)dx$

2. slaaibak

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: $-{1 \over 2} \cos (2x - 3) + C$

3. anonymous

i had positive 1/3 not -1/2

4. slaaibak

How did you get a 1/3 ?

5. anonymous

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

6. slaaibak

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

7. anonymous

i meant 2-3x lol sorry!

8. 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.

9. anonymous

yay : )