mdsk.net
当前位置:首页 >> 1加sinx分之1的积分 >>

1加sinx分之1的积分

∫1/(1+sinx)*dx=∫(1-sinx)/cos^2(x)*dx=∫1/cos^2(x)*dx+∫1/cos^2(x)*dcosx=tanx-1/cosx+C

很简单 ∫ dx/(1 + sinx) = ∫ (1 - sinx)/[(1 + sinx)(1 - sinx)] dx= ∫ (1 - sinx)/(1 - sinx) dx = ∫ (1 - sinx)/cosx dx= ∫ (secx - secxtanx) dx= tanx - secx + C

∫1/(1+sinx) dx=∫(1-sinx) / [(1+sinx)(1-sinx)] dx=∫(1-sinx) / (1-sinx) dx=∫(1-sinx) / cosx dx=∫(secx - secxtanx) dx=tanx - secx + C

用万能公式t=tan(x/2),sinx=2t/(1+t),dx=2dt/(1+t)原式=∫2/(1+t)dt=-2/(1+t)+C=-2/(1+tan(x/2))+C

方法一:∫[1/()]dx =2∫{1/[sin(x/2)+cos(x/2)]^2}d(x/2) =2∫{1/[tan(x/2)+1]}^2{1/[cos(x/2)]^2}d(x/2) =2∫{1/[tan(x/2)+1]}^2[tan(x/2)+1] =-2/[1+tan(x/2)]+C.方法二:∫[1/(1+sinx)]dx =∫{1/[1+cos(π/2-x)]}dx =(1/2)∫{1/[cos(π/4-x/2)]^2}dx =-∫{1/[cos(π/4-x/2)]^2}d(π/4

三角函数的不定积分

方法一:∫[1/(1+sinx)]dx=2∫{1/[sin(x/2)+cos(x/2)]^2}d(x/2)=2∫{1/[tan(x/2)+1]}^2{1/[cos(x/2)]^2}d(x/2)=2∫{1/[tan(x/2)+1]}^2[tan(x/2)+1]=-2/[1+tan(x

三角变换后,分别凑微分 过程如下图:

解 ∫1-1/sinxdx =∫1-cscdx =∫1dx-∫cscdx =x-In|cscx-cotx|+c 希望能帮助你,数学辅导团为您解答,不理解请追问,理解请及时采纳!(*^__^*)

∫[sinx/(1+sinx)]dx=x-tanx+1/cosx+C.C为积分常数.解答过程如下:∫[sinx/(1+sinx)]dx =∫[(1+sinx-1)/(1+sinx)]dx =∫dx-∫[1/(1+sinx)]dx =x-∫{(1-sinx)/[1-(sinx)^2]}dx =x-∫[1/(cosx)^2]dx+∫[sinx/(cosx)^2]dx =x-tanx-∫[1/(cosx)^2]d(cosx) =x-tanx+1/cosx+C

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