Question

Hi everyone! I need help figuring out how to solve integrals. These are giving me a hard time. Here's one of them: INT (x)Sin(2x) dx

Answer 1

you have to use integration by parts

Answer 2

\[\int\limits_{}^{}xsin(2x) dx=-\frac{1}{2}xcos(2x)-\int\limits_{}^{}1 \cdot \frac{-1}{2}\cos(2x) dx\]

Answer 3

let u = x and let dv = Sin(2x)
So du = dx and v = -1/2cos(2x)
So INT = uv - Int(vdu)

Answer 4

And you use that to follow myininaya's solution.

Answer 5

\[=-\frac{1}{2}xcos(2x)+\frac{1}{2}\int\limits_{}^{}\cos(2x) dx\]

Answer 6

let me know if you need help integrating cos(2x)

Answer 7

Thanks for the help guys, I really appreciate it. How, for "v" do I get the: "-1/2cos" ?

Answer 8

\[\int\limits_{}^{}\sin(2x)dx=\frac{-1}{2}\cos(2x)+C \]
we just ignore C until later though

Answer 9

maybe it will be easier for you to see if i do this:
let u=2x =>du=2 dx
so we have
\[\int\limits_{}^{}\frac{1}{2}\sin(u) du=\frac{1}{2}(-\cos(u))+c=-\frac{1}{2}\cos(2x)+c\]

Answer 10

oooh! Okay I see it now! Thanks myininaya!

Answer 11

np