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

Mathematics
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I wrote that incorrectly. -sin(x)+ cos(x) + C
close, but not quite
\[u=x\]\[dv=\cos xdx\]\[\int udv=uv-\int vdu\]look carefully and you should be able to see you dropped an x in there

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Other answers:

@TuringTest , I'm a little confused on your explanation, where have I dropped an x?
u=x, and there is a u on the right of the formula. There should be an x where the u is...
@TuringTest I think I see. So, it would be -sin(x) + x cos(x) + C
what you have above is\[vdu-\int uv\]careful, identify your v and u, dv and du!
\[u=x\implies du=dx\]\[dv=\cos xdx\implies v=-\sin x\]\[\int udv=uv-\int vdu\]slowly and carefully plug in each for u, v, and du
@TuringTest Thank you for your patience with me. Would it be x sin x + cos x + C?
true
\[u=x\implies du=dx\]\[dv=\cos xdx\implies v=-\sin x\]\[\int udv=uv-\int vdu\]\[\int x\cos xdx=x(-\sin x)-\int(-\sin x)dx=-x\sin x+\int\sin xdx\]
crap I made up a negative sign
int cosx = sinx not -sinx
\[u=x\implies du=dx\]\[dv=\cos xdx\implies v=\sin x\]\[\int udv=uv-\int vdu\]\[\int x\cos xdx=x(\sin x)-\int(\sin x)dx=x\sin x+\int\sin xdx\]yeah you were right, my bad :P
If only one day I could understand math that way both of you do... thanks for your help!
welcome!

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