Question

Compute ∫ x cos(x)dx.

Would we use Integration by Parts and compute this as - -sin(x) = Cos(x) + C ????

Answer 1

I wrote that incorrectly. -sin(x)+ cos(x) + C

Answer 2

close, but not quite

Answer 3

\[u=x\]\[dv=\cos xdx\]\[\int udv=uv-\int vdu\]look carefully and you should be able to see you dropped an x in there

Answer 4

@TuringTest , I'm a little confused on your explanation, where have I dropped an x?

Answer 5

u=x, and there is a u on the right of the formula. There should be an x where the u is...

Answer 6

@TuringTest I think I see. So, it would be -sin(x) + x cos(x) + C

Answer 7

what you have above is\[vdu-\int uv\]careful, identify your v and u, dv and du!

Answer 8

\[u=x\implies du=dx\]\[dv=\cos xdx\implies v=-\sin x\]\[\int udv=uv-\int vdu\]slowly and carefully plug in each for u, v, and du

Answer 9

@TuringTest Thank you for your patience with me. Would it be x sin x + cos x + C?

Answer 10

true

Answer 11

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

Answer 12

crap I made up a negative sign

Answer 13

int cosx = sinx not -sinx

Answer 14

\[u=x\implies du=dx\]\[dv=\cos xdx\implies v=\sin x\]\[\int udv=uv-\int vdu\]\[\int x\cos xdx=x(\sin x)-\int(\sin x)dx=x\sin x+\int\sin xdx\]yeah you were right, my bad :P

Answer 15

If only one day I could understand math that way both of you do... thanks for your help!

Answer 16

welcome!