2007-01-07 04:47:35 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
-(x*Cos[x]) + Sin[x]
2007-01-07 04:54:45 · answer #1 · answered by Raoof 2 · 0⤊ 0⤋
you ought to break (xsinx dx) into u and dv. enable (x dx) be u and enable (sinx) be dv. Now considering that u=x dx, then du=dx, and because dv=sinx, then v=-cosx Now use your formula udv=uv-Int(vdu) that's Int(xsinx dx) = x(-cosx)-Int(-cosx dx), take the unfavourable out to circulate away -xcosx+Int(cosx dx) which equals: -xcosx+sinx+C that's your answer, do no longer overlook the consistent C
2016-12-12 06:10:24 · answer #2 · answered by cheng 4 · 0⤊ 0⤋
Yay, integration by parts! I haven't done this in ... 16 years.
u = x
du = dx
dv = sinx dx
v = -cosx dx
∫ u dv = uv - ∫ v du
∫ x sinx dx = -x cosx - ∫ -cosx dx
= -x cosx + sinx + C
Not bad for an old pharte, eh?
2007-01-07 04:52:15 · answer #3 · answered by Jim Burnell 6 · 1⤊ 0⤋