求解两道数学微积分题，急！

Question

如题所述

Answer 1

1、∫(0,π/2) (sinx)^m*(cosx)^n dx
=∫(0,π/2) (sinx)^m*(cosx)^(n-1) d(sinx)
=(sinx)^(m+1)*(cosx)^(n-1)|(0,π/2)-∫(0,π/2) sinx d[(sinx)^m*(cosx)^(n-1)]
=-∫(0,π/2) sinx*[m*(sinx)^(m-1)*(cosx)^n-(n-1)*(sinx)^(m+1)*(cosx)^(n-2)] dx
=(n-1)*∫(0,π/2) (sinx)^(m+2)*(cosx)^(n-2) dx-m*∫(0,π/2) (sinx)^m*(cosx)^n dx
=(n-1)*∫(0,π/2) (sinx)^m*[1-(cosx)^2]*(cosx)^(n-2) dx-m*∫(0,π/2) (sinx)^m*(cosx)^n dx
=(n-1)*∫(0,π/2) (sinx)^m*(cosx)^(n-2) dx-(m+n-1)*∫(0,π/2) (sinx)^m*(cosx)^n dx
所以∫(0,π/2) (sinx)^m*(cosx)^n dx
=[(n-1)/(m+n)]*∫(0,π/2) (sinx)^m*(cosx)^(n-2) dx
根据上述递推公式，∫(0,π/2) (sinx)^m*(cosx)^n dx
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*∫(0,π/2) (sinx)^m*(cosx)^(n-4) dx
=......
（1）若n为奇数，则∫(0,π/2) (sinx)^m*(cosx)^n dx
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*...*[2/(m+3)]*∫(0,π/2) (sinx)^m*cosx dx
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*...*[2/(m+3)]*∫(0,π/2) (sinx)^m d(sinx)
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*...*[2/(m+3)]*[1/(m+1)]*(sinx)^(m+1)|(0,π/2)
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*...*[2/(m+3)]*[1/(m+1)]
=(n-1)!!/[(m+n)!!/(m-1)!!]
=[(m-1)!!*(n-1)!!]/(m+n)!!
（2）若n为偶数，则∫(0,π/2) (sinx)^m*(cosx)^n dx
=[(n-1)/(m+n)]*[(n-3)/(m+n-2)]*...*[1/(m+2)]*∫(0,π/2) (sinx)^m dx
=(n-1)!!/[(m+n)!!/m!!]*∫(0,π/2) (sinx)^m dx
=[m!!*(n-1)!!]/(m+n)!!*∫(0,π/2) (sinx)^m dx
其中，∫(0,π/2) (sinx)^m dx
=-∫(0,π/2) (sinx)^(m-1) d(cosx)
=-(sinx)^(m-1)*cosx|(0,π/2)+(m-1)*∫(0,π/2) (sinx)^(m-2)*(cosx)^2 dx
=(m-1)*∫(0,π/2) (sinx)^(m-2)*[1-(sinx)^2] dx
=(m-1)*∫(0,π/2) (sinx)^(m-2) dx-(m-1)*∫(0,π/2) (sinx)^m dx
所以∫(0,π/2) (sinx)^m dx=[(m-1)/m]*∫(0,π/2) (sinx)^(m-2) dx
=[(m-1)/m]*[(m-3)/(m-2)]*∫(0,π/2) (sinx)^(m-4) dx
=......
①若m为奇数，则∫(0,π/2) (sinx)^m dx
=[(m-1)/m]*[(m-3)/(m-2)]*...*(2/3)*∫(0,π/2) sinx dx
=[(m-1)/m]*[(m-3)/(m-2)]*...*(2/3)*(-cosx)|(0,π/2)
=(m-1)!!/m!!
则∫(0,π/2) (sinx)^m*(cosx)^n dx
=[m!!*(n-1)!!]/(m+n)!!*[(m-1)!!/m!!]
=[(m-1)!!*(n-1)!!]/(m+n)!!
②若m为偶数，则∫(0,π/2) (sinx)^m dx
=[(m-1)/m]*[(m-3)/(m-2)]*...*(1/2)*∫(0,π/2) dx
=[(m-1)!!/m!!]*(π/2)
则∫(0,π/2) (sinx)^m*(cosx)^n dx
=[m!!*(n-1)!!]/(m+n)!!*[(m-1)!!/m!!]*(π/2)
=[(m-1)!!*(n-1)!!]/(m+n)!!*(π/2)
综上所述，
当n为奇数，或者n为偶数且m为奇数时，原式=[(m-1)!!*(n-1)!!]/(m+n)!!
当n为偶数且m为偶数时，原式=[(m-1)!!*(n-1)!!]/(m+n)!!*(π/2)
2、（1）先计算∫(0,π/2) sin(nx)/sinx dx的值
因为sin(nx)
=sin[(n-1)x+x]
=sin(n-1)x*cosx+cos(n-1)x*sinx
=(1/2)*[sin(nx)+sin(n-2)x]+cos(n-1)x*sinx
所以(1/2)*sin(nx)=(1/2)*sin(n-2)x+cos(n-1)x*sinx
sin(nx)=sin(n-2)x+2cos(n-1)x*sinx
sin(nx)/sinx=sin(n-2)x/sinx+2cos(n-1)x
∫(0,π/2) sin(nx)/sinx dx
=∫(0,π/2) sin(n-2)x/sinx dx+∫(0,π/2) 2cos(n-1)x dx
=∫(0,π/2) sin(n-2)x/sinx dx+[2/(n-1)]*sin(n-1)x|(0,π/2)
=∫(0,π/2) sin(n-2)x/sinx dx+[2/(n-1)]*sin[(n-1)π/2]
=∫(0,π/2) sin(n-4)x/sinx dx+[2/(n-3)]*sin[(n-3)π/2]+[2/(n-1)]*sin[(n-1)π/2]
=......
当n是奇数时，∫(0,π/2) sin(nx)/sinx dx
=∫(0,π/2) sinx/sinx dx+(2/2)*sin(2π/2)+(2/4)*sin(4π/2)+...+[2/(n-1)]*sin[(n-1)π/2]
=π/2
（2）再来计算∫(0,π/2) [sin(nx)/sinx]^2 dx
[sin(nx)/sinx]^2
=[sin(n-2)x/sinx+2cos(n-1)x]^2
=[sin(n-2)x/sinx]^2+4sin(n-2)x*cos(n-1)x/sinx+4[cos(n-1)x]^2
=[sin(n-2)x/sinx]^2+2[sin(2n-3)x-sinx]/sinx+2[1+cos(2n-2)x]
=[sin(n-2)x/sinx]^2+2sin(2n-3)x/sinx+2cos(2n-2)x
∫(0,π/2) [sin(nx)/sinx]^2 dx
=∫(0,π/2) [sin(n-2)x/sinx]^2 dx+∫(0,π/2) 2sin(2n-3)x/sinx dx+∫(0,π/2) 2cos(2n-2)x dx
因为2n-3时奇数，所以∫(0,π/2) 2sin(2n-3)x/sinx dx=2*(π/2)=π
∫(0,π/2) [sin(nx)/sinx]^2 dx
=∫(0,π/2) [sin(n-2)x/sinx]^2 dx+π+[1/(n-1)]*sin(2n-2)x|(0,π/2)
=∫(0,π/2) [sin(n-2)x/sinx]^2 dx+π
=∫(0,π/2) [sin(n-4)x/sinx]^2 dx+2π
=......
若n是奇数，则∫(0,π/2) [sin(nx)/sinx]^2 dx
=∫(0,π/2) (sinx/sinx)^2 dx+[(n-1)/2]*π
=π/2+[(n-1)/2]*π
=nπ/2
若n是偶数，则∫(0,π/2) [sin(nx)/sinx]^2 dx
=∫(0,π/2) [sin(2x)/sinx]^2 dx+[(n-2)/2]*π
=∫(0,π/2) 4(cosx)^2dx+[(n-2)/2]*π
=∫(0,π/2) 2[1+cos(2x)]dx+[(n-2)/2]*π
=[2x+sin(2x)]|(0,π/2)+[(n-2)/2]*π
=π+[(n-2)/2]*π
=nπ/2
综上所述，当n为正整数时，∫(0,π/2) [sin(nx)/sinx]^2 dx=nπ/2

Answer 2

第一题的答案：n/（m+1）（m+n+1）