$\displaystyle \displaystyle\int(w^2-6w)sin(2w)dw$

u= $\displaystyle (w^2-6w)$ du= $\displaystyle (2w-6) dw$

dv= $\displaystyle sin(2w)dw$ and v= $\displaystyle \frac{-1}{2}cos(2w)$

$\displaystyle \displaystyle\frac{-1}{2}(w^2-6w)sin(2w)-\int \frac{-1}{2}(2w-6)cos(2w)dw$

u= $\displaystyle (2w-6)$ du= $\displaystyle 2 dw$

dv= $\displaystyle cos(2w)dw$ and v= $\displaystyle \frac{1}{2}sin(2w)$

$\displaystyle \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\frac{1}{2}sin(2w)*2dw$

$\displaystyle \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)-\int\sin(2w)dw$

$\displaystyle \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{2}cos(2w)+c$

Is my algebra wrong? Cause the book is saying that the answer is:

$\displaystyle \displaystyle\frac{-1}{2}(w^2-6w)cos(2w)+\frac{1}{4}(2w-6)sin(2w)+\frac{1}{4}cos(2w)+c$