大一高数，求大神速解

如图

原式=(1/2)∫【0,π/2】sin⁴xsin2xdsinx
= (1/2)∫【0,π/2】sin³xsinxsin2xcosxdx
=(1/4)∫【0,π/2】sin³xsin²2xdx
= (1/16)∫【0,π/2】sinx(cos3x-cosx)²dx
= (1/16)∫【0,π/2】sinx(cos²3x-2cos3 xcosx+cos²x)dx
= (1/16)∫【0,π/2】sinxcos²3 x-sin2xcos3x+(1/2)sin2xcos xdx
= (1/16)∫【0,π/2】(1/2)sinx+(1/4)sin7 x-(1/4)sin5 x-(1/2)sin5 x+(1/2)sinx+(1/4)sin3 x+(1/4)sinxdx
= (1/64)∫【0,π/2】sin7 x-3sin5x+sin3x+5sinx dx
=(1/64)(-cos7x/7+3cos5x/5-cos3x/3-5cosx)【0,π/2】
=(1/64)(2/7-3/5+1/3+5)
=(1/64)(525+30-63+35)/105
=527/6720

注意变换积分上下限。并且最后不要忘记把变量换回来