如图所示，这是由对称性决定的
f(x)=[sin(x)]^4的周期是π，对称轴是x=kπ/2（k为整数）。由对称性、定积分的几何性质知原式成立
(sinx)^2=(1-cos2x)/2，因此(sinx)^2的周期与cos2x相同，等于π
(sinx)^4=[(sinx)^2]^2=[(1-cos2x)/2]^2=(1-cos2x)^2/4=[1-2cos2x+(cos2x)^2]/4=[1-2cos2x+(1+cos4x)/2]/4，(sinx)^4的周期是cos2x的周期（等于π）和cos4x的周期（等于π/2）的最小公倍数，故(sinx)^4的周期是π
以此类推，(sinx)^(2k)=a + b*cos2x + c*cos4x + d*cos6x + ...（k=1,2,3...），周期是π、π/2、π/3……的最小公倍数，即(sinx)^(2k)的周期是π
而(sinx)^(2k)的对称轴是x=kπ/2（k为整数），即在[0,π]内的图形关于x=π/2对称，故有∫(0→π/2)(sinx)^(2k)dx=∫(π/2→π)(sinx)^(2k)dx=(1/2)∫(0→π)(sinx)^(2k)dx
由此推出∫(0→2π)(sinx)^4*dx=2∫(0→π)(sinx)^4*dx=2*2∫(0→π/2)(sinx)^4*dx=4∫(0→π/2)(sinx)^4*dx
(1)
(cosx)^2 = 1- (sinx)^2
I(2n)
=∫(0->π/2) (sinx)^(2n) dx
=-∫(0->π/2) (sinx)^(2n-1) dcosx
= - [cosx.(sinx)^(2n-1)]|(0->π/2) +(2n-1)∫(0->π/2) (sinx)^(2n-2) .(cosx )^2 dx
=0 +(2n-1)∫(0->π/2) (sinx)^(2n-2) .(cosx )^2 dx
=(2n-1)∫(0->π/2) (sinx)^(2n-2) .[1-(sinx )^2] dx
2nI(2n) =(2n-1)I(2n-2)
I(2n) =[(2n-1)/(2n)]I(2n-2)
8∫0->π/2) (sinx)^4 . (cosx)^2 dx + π
=8∫0->π/2) (sinx)^4 .[ 1- (sinx)^2] dx + π
=8[ I4 - I6] +π
=8[ (3/4)(1/2)I0 - (5/6)(3/4)(1/2)I0 ] +π
=8[ (3/4)(1/2)(π/2) - (5/6)(3/4)(1/2)(π/2) ]+π
=4π[ (3/4)(1/2)(1/6) ]+π
=π/4 + π
=5π/4
