早教吧作业答案频道 -->数学-->
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
题目详情
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
▼优质解答
答案和解析
∫[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
=∫[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
看了 求定积分∫(sinx)^(n...的网友还看了以下:
50分送上!利用和差角公式化简:1.3√15sinx+3√5cosx2.√2(sinx-cosx) 2020-04-11 …
求一个简单三角函数化简求一个三角函数化简a×cosβt+b×sinβt化简成A×cos(βt+φ) 2020-06-06 …
高数积分区间讨论问题0到pai上定积分,cos^2(x)dx,cosx在二分之派到派应该是负值,该 2020-07-23 …
用matlab求定积分的上限b的值定积分l=∫(x^2+y^2+z^2)^(1/2)*xd(t)上 2020-07-24 …
帮我约分和解决函数题目谢谢.数学来的如题,数学约分cos^2A=16/25这已经是不是最简?最后怎 2020-07-30 …
用诱导公式求下列三角函数值cos(-17派/4)sin(-1574度)sin(-26派/3)sin 2020-08-02 …
数学救急啊200分cos(75度+阿尔法)=三分之一且阿尔法大于负180度小于负90度则cos(1 2020-08-02 …
重金求解积分:∫(cosθ/(a*cosθ+b))dθ=?从0到2π的积分值a,b为常数.∫(co 2020-08-02 …
已知θ是第二象限角,求cosθ/2=-1/2,那么√(1-sin2θ)/(cosθ/2-sinθ/ 2020-08-03 …
求积分cos^n(x)*cos(nx)求此式子积分(cos(x))^n*cos(nx) 2020-11-07 …