求∫sin^2xcos^2xdx的不定积分,详细过程 - : ∫ xcos^2xdx =∫ x(1+cos2x/2)dx =1/2∫ xdx+1/2∫xcos2xdx =x²/4+1/4∫xdsin2x =x²/4+1/4*xsin2x-1/4∫sin2xdx =x²/4+1/4*xsin2x-1/8∫sin2xd2x =x²/4+1/4*xsin2x+1/8*cos2x+c
∫sin^2xcos^2xdx - : ∫sin^2xcos^4xdx=∫(1-cos^2x)cos^4xdx=∫cos^4xdx-∫cos^6xdx 利用积分表得到.
跪求∫cos2x/sin^2xcos^2xdx 十万火急!! - : sin^2xcos^2x=1/4(2sinxcosx)^2=1/4sin2x 原式=∫cos2x/sin^2xcos^2xdx=∫cos2x/(1/4sin2x)dx=∫2/sin2xd(sin2x)=2lnsin2x+C
sin^2xcos^2x不定积分 - : ∫sin^2xcos^2xdx=1/4∫(sin2x)^2dx=1/8∫(1-cos4x)dx=1/8(x-1/4sin4x)+C=x/8-(sin4x)/32+C满意请采纳,
积分符号1/sin^2xcos^2xdx - : =∫x(1+cos2x)/2dx=x^2/4+∫xcos2x/4d(2x)=x^2/4+∫x/4d(sin2x)=x^2/4+(xsin2x)/4-1/4∫(sin2x)dx=x^2/4+(xsin2x)/4-1/8∫(sin2x)d(2x)=x^2/4+(xsin2x)/4+1/8cos(2x)+C
跪求大神解答高数∫sin^2xcos^2*3xdx - : ∫sin^版2xcos^权3xdx =∫(sinx)^2cosx(1-(sinx)^2)dx =∫(sinx)^2cosxdx-∫(sinx)^4cosxdx =∫(sinx)^2d(sinx)-∫(sinx)^4d(sinx) =(sinx)^3/3-(sinx)^5/5+C