## bmelyk Group Title Integrate the following: one year ago one year ago

1. bmelyk Group Title

$∫ \sin(2x-3)dx$

2. slaaibak Group Title

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. bmelyk Group Title

i had positive 1/3 not -1/2

4. slaaibak Group Title

How did you get a 1/3 ?

5. bmelyk Group Title

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

6. slaaibak Group Title

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

7. bmelyk Group Title

i meant 2-3x lol sorry!

8. hba Group Title

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. bmelyk Group Title

yay : )