∫（xsin x）²dx 不定积分怎么求

如题所述

第1个回答 2011-12-20

∫（xsin x）²dx
=Sx^2*(sinx)^2 dx
=Sx^2*(1-cos2x)/2 dx
=1/2*Sx^2dx-1/2*Sx^2 cos2x dx
=1/6*x^3-1/4*Sx^2dsin2x
=1/6*x^3-1/4*x^2sin2x+1/4*Ssin2xdx^2
=1/6*x^3-1/4*x^2sin2x+1/2*Sxsin2xdx
=1/6*x^3-1/4*x^2sin2x-1/4*Sxdcos2x
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/4*Scos2xdx
=1/6*x^3-1/4*x^2sin2x-1/4*xcos2x+1/8*sin2x+c本回答被提问者采纳

相似回答

求不定积分∫xsin²xdx 在线等答：∫xsin²xdx=(1/2)∫x(1-cos2x)dx=(1/2)∫xdx-(1/2)∫ xcos2xdx =(1/4)x^2-(1/4)∫ xd(sin2x)=(1/4)x^2-(1/4)xsin2x+(1/4)∫ (sin2x)dx =(1/4)x^2-(1/4)xsin2x-(1/8)cos2x+C

高数求不定积分答：∫xsin²xdx =1/2*∫x(1-cos2x)dx =1/2*∫xdx-1/2*∫xcos2xdx =x²/4-1/4*∫xdsin2x =x²/4-1/4*(xsin2x-∫sin2xdx)=x²/4-1/4*(xsin2x+1/2*∫dcos2x)=x²/4-xsin2x/4-cos2x/8 ...

高等数学求不定积分答：∫xsin²xdx =∫x[1-cos(2x)]/2 dx =½∫[x-xcos(2x)]dx =½∫xdx -½∫xcos(2x)dx =½∫xdx -¼∫xd[sin(2x)]=¼x²-¼xsin(2x)+¼∫sin(2x)dx =¼x²-¼xsin(2x)-⅛cos(2x)+C ...

高数 不定积分 求大神答：∫ 1/cos²xsin²x dx =∫ (cos²x+sin²x)/cos²xsin²x dx =∫ (1/cos²x+1/sin²x)dx =tanx-cotx+C

求不定积分∫xsin(x+1)dx答：回答：使用分部积分即可

求不定积分∫x*(sin(x))^2,要过程,谢谢答：解：∫xsin²xdx =∫x(1-cos2x)/2 dx =∫x/2dx-∫(xcos2x)/2dx =x²/4-[x(sin2x)/4-∫(sin2x)/4 dx]=x²/4-[x(sin2x)/4+(cos2x)/8]+C =x²/4-x(sin2x)/4-(cos2x)/8+C

求不定积分∫xsin;答：方法如下，请作参考：若有帮助，请采纳。

大家正在搜