changing signs in integration by part

[legacyofpiracy]

Sorry I keep bombarding this board with questions, just trying to make sense of all of this. I have been having a bit of trouble in solving integration by parts, particularily dealing with the signs after you plug your values into the equation " S udv= uv - S vdu " here is an example:

 Use integration by parts to evaluate the integral 

S 4xsinxdx

u=4x      du=4xdx
v=-cosx  dv=sinxdx

-cosx4x - S -cosx4dx

4x(-cosx)+4sinx+C

This final answer is correct according to my book, however I am confused as to how the + sign comes into play. When integrating the -cos it becomes a positive sign, leaving behind another negative sign before the integegral sign ("S"). How does it become positive then? Let me know if you need clarification, i'm not sure how well I did at explaining my problem...

 

[Unco]

\(\displaystyle \begin{align*}
\mbox{ - \int -4\cos{(x)} dx }&=\mbox{ -(-4) \int \cos{(x)} dx }\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&=\mbox{ 4 \int \cos{(x)} dx }\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
&=\mbox{ 4\sin{(x)} + C}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\
\end{align*}\)


To re-iterate: \(\displaystyle \mbox{\int -\cos{(x)} dx = -\int \cos{(x)} dx = - \sin{(x)} + C}\)

 

[legacyofpiracy]

ooh! that makes alot more sense..thanks alot!