knrt.net
当前位置:首页 >> 求极限limx→0+∫2x0|x?t|sintDtx3 >>

求极限limx→0+∫2x0|x?t|sintDtx3

解 limx→0+∫2x0|x?t|sintdtx3=limx→0+∫x0(x?t)sintdt+∫2xx(t?x)sintdtx3=limx→0+∫x0sintdt+2xsin2x?∫2xxsintdt3x2=limx→0+sinx6x+43?limx→0+2sin2x?sinx6x=16+43?46+16=1.

令x=tant,dx=(sect)2dt,原式=∫(tant)3+1/(sect)2dt=∫tant(sint)2dt+∫(cost)2dt=∫tantdt-∫costsintdt+t/2+sin2t/4+c=-ln|cost|-∫sin2td(2t)/4+t/2+sin2t/4+c=-ln1/√1+x2+(cos2t+sin2t)/4+(arctanx)/2+c=(ln(1+x2)+(x+1)/...

利用级数可以做吧, tanx=x+x^3/3+2x^5/15+O(x^6)=T+O(x^6), tanT=T+T^3/3+2T^5/15+O(T^6)=x+2x^3/3+3x^5/5+O(x^6); sinx=x-x^3/6+x^5/120+O(x^6)=S+O(x^6), sinS=S-S^3/6+S^5/120+O(S^6)=x-x^3/3+x^5/10+O(x^6). 则 lim[tan(tanx)-sin(sinx)]/x...

网站首页 | 网站地图
All rights reserved Powered by www.knrt.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com