Question

.

Answer 1

what subs have you tried?

Answer 2

wait, is this\[\int-6x\cos(2x)dx\]?

Answer 3

substitution will not work, you will need integration by parts
do you know how to do that?
-oh, that's what you meant...

Answer 4

so what have you tried for your substitutions?
u=?
dv=?

Answer 5

v is incorrect

Answer 6

now u du is incorrect as well
you were right the first time
du=-6

Answer 7

you do carry dx to the other side, not x\[u=-6x\implies du=-6dx\](it's not really a derivative, it's a differential; remember those from calc I?)

Answer 8

good :)
to find v we need\[v=\int dv=\int\cos(2x)dx\]which requires a u-substitution
try that integral again and tell me what v is ;)

Answer 9

\[v=\frac12\sin(2x)\]

Answer 10

no wait, you got that wrong

Answer 11

the formula for integration by parts is\[\int udv=uv-\int vdu\]try plugging in what we found for u,v, and du again

Answer 12

\[dv=\cos(2x)\implies v=\frac12\sin(2x)\]\[u=-6x\implies du=-6dx\]\[\int udv=uv-\int vdu\]\[\int-6x\cos(2x)dx=-3\sin(2x)+3\int\sin(2x)dx\]

Answer 13

um...
I'm not sure what you are seeing
right-click the math font that you cannot read and go to "math settings"
there you can try to mess with the zoom or the render

Answer 14

I hope you can see it now
and it's fine, I actually learned something by messing with it myself :)
anyway, yeah
integrate and add C

Answer 15

you're welcome