mdsk.net
当前位置:首页 >> ∫1/(1+sin^2x)Dx定积分 上限是3π/4,下线是0。自己求出来有问题。求详细解答 >>

∫1/(1+sin^2x)Dx定积分 上限是3π/4,下线是0。自己求出来有问题。求详细解答

这题的不定积分过程应该没有困难,我想你的问题在于最后代入积分限时出错.注意:原函数在x=π/2处是个间断点:那么就需要分区间代入积分结果,因为牛顿-莱布尼兹公式要求区间上函数是连续的,参考下图:

我想你的疑问应该也在这里:lim(x→0) tanx = lim(x→π) tanx = 0 ?x = π/2是tanx的间断点,所以应该分为[0,π/2)U(π/2,π]

令tanx = t,x = arctant 则dx = dt/(1+t)1+sinx = 1 + t/(1+t) ∫dx/(1+sinx)=∫dt/(1+2t)=1/√2 ∫d√2t/[1+(√2t)]=1/√2 arctan√2t + C=1/√2 arctan(√2tanx) + C x = π/2时,1/√2 arctan(√2tanx) = π/2√2 x = 0时,1/√2 arctan(√2tanx) = 0 原式 = π/2√2

解:设tanx=t,则x=arctant,sinx=t/√(t+1),dx=dt/(t+1) 于是,原式=∫[dt/(t+1)]/[1+t/(t+1)] =∫dt/(2t+1) =(1/√2)∫d(√2t)/[(√2t)+1] =(1/√2)arctan(√2t)+C (C是积分常数) =(1/√2)arctan(√2tanx)+C.

原式=∫x*csc^2x dx(下限π/4,上限π/3)=-(1/2)*∫xd(cot2x)(下限π/4,上限π/3)=-(1/2)*xcot2x+(1/2)*∫cot2xdx(下限π/4,上限π/3)=-(1/2)*(π/3)*cot(2π/3)+(1/4)*∫(cos2x/sin2x)d(2x)(下限π/4,上限π/3)=√3π/18+(1/4)*∫d(sin2x)/sin2x(下限π/4,上限π/3)=√3π/18+(1/4)*ln|sin2x|(下限π/4,上限π/3)=√3π/18+(1/4)*ln(√3/2)=√3π/18+(ln3)/8-(ln2)/4

解:遇到三角函数考虑万能公式.∫x/(sinx)^2dx=∫x/[(1-cos2x)/2]dx=∫2x/(1-cos2x)dx=∫2x/[1-(1-tgxtgx)/(1+tgxtgx)]dx=∫2x(1+tgxtgx)/(2tgxtgx)dx=∫[x/(tgxtgx)]d(tgx) (tgx)'=(1+tgxtgx)=-x/tgx +∫(1/tgx)dx =-x/tgx +lnsinx +C [π/4,π/3]代入:原式=-π/(3sqr3)+0.5(ln3-3ln2)+π/4 ( sqrA是根号A)

∫1/(2-(sin2x))dx分子分母同除以(cos2x),得=∫sec2x/(2sec2x-tan2x)dx=1/2∫sec2x/(2sec2x-tan2x)d(2x)=1/2∫1/(2sec2x-tan2x)d(tan2x)=1/2∫1/(2(1+tan2x)-tan2x)d(tan2x)=1/2∫1/(2+tan2x)d(tan2x)=(1/2)*(1/√2)arctan((tan2x)/√2)+C=(√2/4)arctan((tan2x)/√2)+C 代入上下限结果为:√2π/8

计算过程如下:∫ 1/(1+sin^2x)dx= ∫ [1/cos^2x]/(1/cos^2x+tan^2x)dx= ∫ [sec^2x]/(sec^2x + tan^2x)dx= ∫ 1/(1 + 2tan^2x)dtanx= 1/√2 *∫ 1/(1 + (√2tanx)^2)d(√2tanx)= 1/√2 * arctan(√2tanx) + C(C为常数) 扩展资料:不定积分求法:1、积分公式法.

=(1/2)∫dx/sin2x=(1/4)ln|cot2x-cot2x|+C 代入上下限即可

楼主,注意下素质.那个上下限你自己知道就行,我不好写出来. ∫(1-sin2x)/(1+sin2x) dx =∫[(cosx -sinx)/(cosx+sinx)]^2 dx =∫tan(x -π/4) *tan(x-π/4) dx ① 注意到 (tan(x-π/4))' =1+tan(x-π/4)*tan(x-π/4) ∴ ①可以化为 ∫tan(x -π/4) *tan(x-π/4) dx =∫[ 1+tan(x-π/4)*tan(x-π/4)] dx -∫dx =tan(x-π/4)|(这里写上下限0,π) -(π-0) =(tan(π-π/4) -tan(0-π/4)) -π = -π

相关文档
ceqiong.net | wnlt.net | wlbx.net | xmjp.net | ldyk.net | 网站首页 | 网站地图
All rights reserved Powered by www.mdsk.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com