知道答案往回推我也会呀😊请顺着推下来好吗,或者说有什么公式
追答这个,就是公式啊。。楼主明鉴。
要推,就按楼主的思路。
x = sec(t), 0 1
原式=Ssec(t)tan(t)dt/|tan(t)|
=Ssec(t)dt = Sdt/cos(t) = Scos(t)dt/[cos(t)]^2 = Sd[sin(t)]/{1 - [sin(t)]^2}
= (1/2)Sd[sin(t)]/[1-sin(t)] + (1/2)Sd(sin(t))/[1 + sin(t)]
= (-1/2)ln[1-sin(t)] + (1/2)ln[1+sin(t)] + C
= (1/2)ln{[1+sin(t)]/[1-sin(t)]} + C
= (1/2)ln{ [1 + (1-1/x^2)^(1/2)]/[ 1 - (1-1/x^2)^(1/2)] } + C
= (1/2)ln{[ x + (x^2 - 1)^(1/2)]/[ x - (x^2 - 1)^(1/2)] } + C
= (1/2)ln{ [x + (x^2 - 1)^(1/2)]^2/ [x^2 - (x^2 - 1)] } + C
= ln[ x + (x^2 - 1)^(1/2)] + C.
PI<t<PI..超字数了。。
再问,再贴。。好吗?
或令2x-1=sect推导
貌似是考研数学哎