定积分计算题,求解答过程,望诸大神帮给个解答,不胜感激!(其实就是同济第六版高数课本上的例题中的最 几道高数题,求大神解答,给出具体过程?
\u4e00\u9053\u9ad8\u6570\u9898\uff0c\u6c42\u89e3\u7b54!\u8be6\u7ec6\u5b8c\u6574\u6e05\u695a\u8fc7\u7a0brt\u6240\u793a\u2026\u5e0c\u671b\u80fd\u5e2e\u5230\u4f60
\u671b\u91c7\u7eb3\uff0c\u5e0c\u671b\u53ef\u4ee5\u5e2e\u52a9\u5230\u4f60\u3002\u6700\u540e\u4e00\u4e2a\u9898\u6709\u4e0d\u61c2\u7684\u5730\u65b9\u53ef\u4ee5\u518d\u95ee\u6211\u3002
定积分不好书写,给你一个不定积分的过程
令θ=tanα,则:√(1+θ^2)=√[1+(tanα)^2]=1/cosα, dθ=[1/(cosα)^2]dα。
sinα=√{(sinα)^2/[(sinα)^2+(cosα)^2]}=√{(tanα)^2/[1+(tanα)^2}
=θ/√(1+θ^2),
∴原式=∫{(1/cosα)[1/(cosα)^2]}dα
=∫[cosα/(cosα)^4]dα
=∫{1/[1-(sinα)^2]^2}d(sinα)。
再令sinα=u,则:
原式=∫[1/(1-u^2)^2]du
=(1/4)∫[(1+u+1-u)^2/(1-u^2)^2]du
=(1/4)∫[(1+u)^2/(1-u^2)^2]du+(1/2)∫[(1-u^2)/(1-u^2)^2]du
+(1/4)∫[(1-u)^2/(1-u^2)^2]du
=(1/4)∫[1/(1-u)^2]du+(1/2)∫[1/(1-u^2)]du+(1/4)∫[1/(1+u)^2]du
=-(1/4)∫[1/(1-u)^2]d(1-u)+(1/4)∫[(1+u+1-u)/(1-u^2)]du
+(1/4)∫[1/(1+u)^2]d(1+u)
=(1/4)[1/(1-u)]-(1/4)[1/(1+u)]+(1/4)∫[1/(1-u)]du
+(1/4)∫[1/(1+u)]du
=(1/4)[1/(1-sinα)]-(1/4)[1/(1+sinα)]
-(1/4)∫[1/(1-u)]d(1-u)+(1/4)∫[1/(1+u)]d(1+u)
=(1/4){1/[1-θ/√(1+θ^2)]}-(1/4){1/[1+θ/√(1+θ^2)]}
-(1/4)ln|1-u|+(1/4)ln|1+u|+C
=(1/4)[1+θ/√(1+θ^2)-1+θ/√(1+θ^2)]/[1-θ^2/(1+θ^2)]
+(1/4)ln|1+sinα|-(1/4)ln|1-sinα|+C
=(1/4)[2θ/√(1+θ^2)]/[(1+θ^2-θ^2)/(1+θ^2)]
+(1/4)ln[|1+θ/√(1+θ^2)|/|1-θ/√(1+θ^2)|]+C
=(1/2)θ√(1+θ^2)+(1/4)ln|[√(1+θ^2)+θ]/[√(1+θ^2)-θ]|+C
=(1/2)θ√(1+θ^2)+(1/4)ln|[√(1+θ^2)+θ]^2/(1+θ^2-θ^2)|+C
=(1/2)θ√(1+θ^2)+(1/2)ln|θ+√(1+θ^2)|+C
8
绛旓細瀹氱Н鍒嗘眰瑙g瓟杩囩▼,澶у楂樼瓑鏁板姘村钩 鎴戞潵绛 1涓洖绛 #璇濋# 鍔冲姩鑺傜函绾庡共璐с,绛変綘鐪!maths_hjxk 2015-04-22 路 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 maths_hjxk 鐭ラ亾鍚堜紮浜烘暀鑲茶瀹 閲囩撼鏁:9802 鑾疯禐鏁:19095 姣曚笟鍘﹂棬澶у姒傜巼璁轰笌鏁扮悊缁熻涓撲笟 纭曞+瀛︿綅 鍚慣A鎻愰棶 绉佷俊TA 鍏虫敞 ...
绛旓細鍑戜釜寰垎灏卞ソ鍋氫簡
绛旓細浣犲ソ锛岃В绛旇繃绋濡備笅锛屽笇鏈涘浣犳湁鎵甯姪锛屾湜閲囩撼锛侊紒锛佽阿璋
绛旓細瀹氱Н鍒鐨璁$畻棰,瑕佹湁璇︾粏瑙g瓟杩囩▼,鏈濂芥墜鍐欐埅鍥,婊℃剰鍔犲垎銆 鎴戞潵绛 1涓洖绛 #鐑# 璇ヤ笉璇ヨ瀛╁瓙寰堟棭瀛︿範浜烘儏涓栨晠?94lmz 2016-02-11 路 TA鑾峰緱瓒呰繃750涓禐 鐭ラ亾灏忔湁寤烘爲绛斾富 鍥炵瓟閲:786 閲囩撼鐜:100% 甯姪鐨勪汉:563涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 鏇村杩介棶杩界瓟 杩介棶 濂藉帀瀹...
绛旓細绗簩灏忛棶鍜岀鍥涘皬闂叿浣瑙g瓟杩囩▼濡備笅鍥炬墍绀:
绛旓細瑙:鍙绉垎鍙橀噺1<=x<=e 闈㈢Н鍏冪礌dA=[(lnx)-(0)]dx 鎵浠 A=鈭玔1,e]lnxdx =xlnx|[1,e]-x|[1,e]=1
绛旓細鍋氬彉鎹 t=鈭(1+e^x) ,x=ln(t^2-1) dx=2tdt/(t^2-1)鈭1/鈭(1+e^x)dx=\鈭(1/t) 2tdt/(t^2-1)=鈭2/(t^2-1)dt =鈭1/(t-1)-1/(t+1)dt =ln(t-1)-ln(t+1)+C =ln(鈭(1+e^x)-1)-ln(鈭(1+e^x)+1)+C 濡傛灉鎴戠殑鍥炵瓟瀵逛綘鏈夊府鍔﹡璇风偣鍑汇愭垜...
绛旓細鍥炵瓟锛氳缁瑙g瓟杩囩▼濡備笅鍥剧墖: 璇峰湪姝よ緭鍏ユ偍鐨勫洖绛
绛旓細娌℃湁鍒濈瓑鍘熷嚱鏁般
绛旓細宸茬煡v锛18km/h锛5m/s锛宼锛11min50s锛710s锛孡锛280m 瑙:S锛漹t鈶 S锛堟ˉ锛夛紳S锛婰鈶 鐢扁憼鈶¤В寰 S锛堟ˉ锛夛紳3830m 绛:杩欏骇澶фˉ鐨勯暱搴︿负3830m銆