作业帮求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法
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求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法
求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法
求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法
∫[0,π] sinx^(n-1) cosx^(n+1)dx
=∫[0,π]sinx^(n-1)cosx^(n-1)*cosx^2dx
=(1/2^n)∫[0,π](sin2x)^n [(1+cos2x)/2 ]dx
= (1/2^n)∫[0,π]sin(2x)^ndx - (1/2^(n+2))∫[0,π]sin(2x)^ndsin2x
=(1/2^(n+1))∫[0,π]sin(2x)^(n-1)dcos2x -(1/2^(n+2))(1/(n+1))(sin2x)^(n+1))|[0,π]
=(1/2^(n+1))(sin2x)^(n-1)cos2x|[0,π] -(1/2)^(n+1)∫[0,π]cos2xd(sin2x)^(n-1)
=(-1/2^n+1)∫[0,π] (n-1) (sin2x)^(n-2)(cos2x)^2d(2x)
=(-1/2^(n+1)∫[0,π](n-1)[(sin2x)^(n-2)-(sin2x)^n] d(2x)
=(-1/2^n)∫[0,π](n-1)(sin2x)^(n-2)dx+(n-1)/2^n ∫[0,π](sin2x)^ndx
In=∫[0,π](sin2x)^ndx
(1/2^n)In= -(n-1)/2^n ∫[0,π](sin2x)^(n-1)dx +(n-1)/2^n∫[0,π] (sin2x)^ndx
nIn=(n-1)In-2
in=(n-1)/n In-2
I1=∫[0,π]sin2xdx=(1/2)cos2x|[0,π]=0
I2=∫[0,π](sin2x)^2dx=(1/4)∫[0,π](1-cos4x)dx =π/2
n偶数时 I4=(3/4)I2=(3/4)(π/2) In=[ 3*5*..(n-1)/(4*6*..*n) ] *(π/2)
∫[0,π] sinx^(n-1)cosx^(n+1)dx=(1/2^n)In=(π/2^(n+1)) [3*5*..*(n-1)/(4*6*..*n)]
n奇数 I3=(2/3)I1=(2/3) In=0
∫[0,π]sinx^(n-1)cosx^(n+1)dx=(1/2^n)In=0