=∫(0，π/2)(sinx)^m*(cosx)^(n- 1)d(sinx)
= (sinx) (m+1) * (cosx) (n-1) | (0,π/2)-∫ (0,π/2) Sinks d [(sinx) m * (cosx).
=-∫(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
So ∫ (0，π/2) (sinx) m * (cosx) n dx
=[(n- 1)/(m+n)]*∫(0，π/2) (sinx)^m*(cosx)^(n-2) dx
According to the above recursive formula, ∫ (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) If n is odd, then ∫ (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) If n is an even number, then ∫ (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
Where ∫ (0, π/2) (sinx) mdx
=-∫(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
So ∫ (0，π/2)(sinx)mdx =[(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) if m is odd, then ∫ (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！ !
Then ∫ (0, π/2) (sinx) m * (cosx) n dx
=[m！ ! *(n- 1)！ ! ]/(m+n)！ ! *[(m- 1)！ ! /m！ ! ]
=[(m- 1)！ ! *(n- 1)！ ! ]/(m+n)！ !
② If m is an even number, then ∫ (0, π/2) (sinx) m dx
=[(m- 1)/m]*[(m-3)/(m-2)]*...*( 1/2)*∫(0，π/2) dx
=[(m- 1)！ ! /m！ ! ]*(π/2)
Then ∫ (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)
All in all,
When n is odd, or n is even and m is odd, the original formula =[(m- 1)! ! *(n- 1)！ ! ]/(m+n)！ !
When n is even and m is even, the original formula =[(m- 1)! ! *(n- 1)！ ! ]/(m+n)！ ! *(π/2)
2.( 1) Calculate the value of ∫(0, π/2) sin(nx)/sinx dx first.
Because of 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
So (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]
= ......
When n is odd, ∫(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) Then calculate ∫ (0, π/2) [sin (NX)/sinx] 2dx.
[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
Because 2n-3 is an odd number, ∫ (0, π/2) 2sin (2n-3) x/sinxdx = 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
= ......
If n is odd, then ∫ (0, π/2) [sin (NX)/sinx] 2dx.
=∫(0，π/2)(sinx/sinx)^2 dx+[(n- 1)/2]*π
=π/2+[(n- 1)/2]*π
=nπ/2
If n is even, then ∫ (0, π/2) [sin (NX)/sinx] 2dx.
=∫(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
To sum up, when n is a positive integer, ∫ (0, π/2) [sin (NX)/sinx] 2dx = nπ/2.