求不定积分∫(x^2+1)sin^2xdx
2013-01-11 17:29:171***
∫(x^2+1)sin^2xdx求不定积分∫(x^2+1)sin^2xdx：设F(x)=∫(x^2+1)sin^2xdx,G(x)=∫(x^2+1)cos^2xdx
F(x)+G(x)=∫(?
最佳回答
设F(x)=∫(x^2+1)sin^2xdx,G(x)=∫(x^2+1)cos^2xdx
F(x)+G(x)=∫(x^2+1)dx=x^3/3+x+C1
G(x)-F(x)=∫(x^2+1)cos(2x)dx
=(1/2)(x^2+1)sin2x-（1/2)∫2xsin2xdx
=(1/2)(x^2+1)sin2x-(1/2)[-(1/2)2xcos2x+(1/2)∫cos2xdx]
=(1/2)(x^2+1)sin2x+(1/2)xcos2x+(1/4)sin2x+C2
两式相减除以2即得（相加除以2可得另一个）
原积分=x^3/6+x/2-(x^2/4)sin2x-(x/4)cos2x-(1/8)sin2x+C
2013-01-17 15:24:22
∫(x^2+1)sin^2(x)dx=(1/2)∫(x^2+1)(1-cos2x)dx
=(1/2)[(x^2+1)(x-(1/2)sin2x)]-∫(x(x-(1/2)sin2x)dx
=(1/2)[(x^2+1)(x-(1/2)sin2x)]-(x^3)/3+(1/2)∫xsin2xdx
=(1/2)[(x^2+1)(x-(1/2)sin2x)]-(x^3)/3-(1/4)xcos2x+(1/4)∫cos2xdx
=(1/2)[(x^2+1)(x-(1/2)sin2x)]-(x^3)/3-(1/4)xcos2x+(1/8)sin2x+C
2013-01-11 18:03:44
答案在文件里
2013-01-11 17:54:53
很赞哦！ (198)
