let y=sin^(2n-1)xcosx
find dy/dx ,
hence show that (2n-1)∫sin^(2n-2)xdx-2n∫sin^(2n)xdx = sin^(2n-1)xcosx+C
A math
2007-12-14 7:10 am
回答 (2)
2007-12-14 7:24 am
✔ 最佳答案
y = sin^(2n - 1)x cosx dy/dx
= (2n - 1) sin^(2n - 2)x cosx cosx + sin^(2n - 1)x (-sinx)
= (2n - 1) sin^(2n - 2)x cos²x - sin^(2n)x
= (2n - 1) sin^(2n - 2)x (1 - sin²x) - sin^(2n)x
= (2n - 1) sin^(2n - 2)x - (2n - 1) sin^(2n)x - sin^(2n)x
= (2n - 1) sin^(2n - 2)x - 2n sin^(2n)x
(2n - 1) sin^(2n - 2)x - 2n sin^(2n)x = d/dx [sin^(2n - 1)x cosx]
(2n - 1) ∫sin^(2n - 2)xdc - 2n ∫sin^(2n)xdc = sin^(2n - 1)x cosx + C
2007-12-14 7:25 am
y=sin^(2n-1)xcosx
dy/dx = (2n-1)sin^(2n-2)x(cosx)(cosx) + sin^(2n-1)x(-sinx)
= (2n-1)sin^(2n-2)x(cos²x) - sin^(2n)x
= (2n-1)sin^(2n-2)x(1-sin²x) - sin^(2n)x
= (2n-1)sin^(2n-2)x-(2n-1)sin^(2n)x - sin^(2n)x
= (2n-1)sin^(2n-2)x-(2n)sin^(2n) x
so dy/dx = (2n-1)sin^(2n-2)x-(2n)sin^(2n) x
intergrating both sides w.r.t x
y + C = ∫(2n-1)sin^(2n-2)x dx-∫(2n)sin^(2n) x dx
sin^(2n-1)xcosx + C = 2n-1∫sin^(2n-2)x dx- 2n∫sin^(2n) x dx
dy/dx = (2n-1)sin^(2n-2)x(cosx)(cosx) + sin^(2n-1)x(-sinx)
= (2n-1)sin^(2n-2)x(cos²x) - sin^(2n)x
= (2n-1)sin^(2n-2)x(1-sin²x) - sin^(2n)x
= (2n-1)sin^(2n-2)x-(2n-1)sin^(2n)x - sin^(2n)x
= (2n-1)sin^(2n-2)x-(2n)sin^(2n) x
so dy/dx = (2n-1)sin^(2n-2)x-(2n)sin^(2n) x
intergrating both sides w.r.t x
y + C = ∫(2n-1)sin^(2n-2)x dx-∫(2n)sin^(2n) x dx
sin^(2n-1)xcosx + C = 2n-1∫sin^(2n-2)x dx- 2n∫sin^(2n) x dx
收錄日期: 2021-04-13 14:42:54
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