∫[sin^2(x)]*[cos^2(x)]dx
=∫（sinxcosx）^2dx
=∫（sin2x/2）^2dx
=1/4∫(sin2x)^2dx
=1/8∫(1-cos4x)dx
=x/8-1/32∫cos4xd4x
=x/8-1/32sin4x+C
希望对您有帮助!
如有不明白,可以追问!
谢谢采纳!
[sin^2(x)]*[cos^2(x)]的不定积分
∫[sin^2(x)]*[cos^2(x)]dx
=∫（sinxcosx）^2dx
=∫（sin2x/2）^2dx
=1/4∫(sin2x)^2dx
=1/8∫(1-cos4x)dx
=x/8-1/32∫cos4xd4x
=x/8-1/32sin4x+C
希望对您有帮助!
如有不明白,可以追问!
谢谢采纳!