integral from 0 to 3 5x Sin (3x+1) I'm pretty sure this is an integration by parts. I made U 5x and dv = sin (3x+1). Is this the correct path?

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integral from 0 to 3 5x Sin (3x+1) I'm pretty sure this is an integration by parts. I made U 5x and dv = sin (3x+1). Is this the correct path?

Mathematics
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\[\int\limits_{0}^{3} 5x \sin(3x+1)dx\] like this?
yes by parts would do it
By parts should work, u = 5x, dv = sin(3x+1) so now integrate for v

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Note: \[\int\limits u dv = uv - \int\limits v du \]
so would v end up being \[\frac{ 1 }{ 3 } \sin (3x+1) \] sorry for the dumb question. Been a couple of years since Calculus 1!
oops, change that sin to a -cos.
Yup
-1/3cos(3x+1)
Keep going, I'm sure you'll get it!

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