## husbishag 2 years ago Compute ∫ x cos(x)dx. Would we use Integration by Parts and compute this as - -sin(x) = Cos(x) + C ????

1. husbishag

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

2. TuringTest

close, but not quite

3. TuringTest

$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

4. husbishag

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

5. TuringTest

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

6. husbishag

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

7. TuringTest

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

8. TuringTest

$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

9. husbishag

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

true

11. TuringTest

$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$

12. TuringTest

crap I made up a negative sign

int cosx = sinx not -sinx

14. TuringTest

$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

15. husbishag

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

16. TuringTest

welcome!