设x=sin(a) a取值范围为[0,pi/2]
（2/3）∫(1-x^2)的3/2次方dx
=（2/3）∫cos^3(a)dsin(a)
=（2/3）∫cos^4(a)da
=（2/3）∫cos^4(a)da
=(2/3)∫[(1+cos2a)/2]^2da
=(2/3)∫(1+cos^2(2a)+2cos(2a))/4da
=(2/3)∫1/4da+(2/3)∫cos^2(2a)/4da+(2/3)∫cos(2a)/2da
=1/6+1/12∫[1+cos(4a)]da+1/3∫cos(2a)da
=1/6+1/12∫da+1/48∫cos(4a)]d4a+1/6∫cos(2a)d2a
=1/6+1/12+(1/48)sin4a|(0,pi/2)+1/6sin2a|（0,pi/2）
=1/6+1/12
=1/4