sec2x的不定积分推导 sec2x的积分证明过程

原式=∫d(tanx)/(1+2tan^2(x) =√2/2∫d(√2tanx)/[1+(√2tanx)^2] =√2/2arctan(√2tanx)+C ‍

令1+tanx=u x=arctan(u-1) dx=du/(1+(u-1)^2)原式=∫du/u(u^2-2u+2)=1/2*∫1/u-(u-2)/(u^2-2u+2)du=1/2*ln|u|-1/2*∫(u-2)du/[(u-2)^2+2(u-2)+2]令u-2=t=1/2*ln|u|-1/2*∫tdt/(t^2+2t+2)=1/.

∫cos(lnx)dx的不定积分为1/2(x*cos(lnx)+x*sin(lnx))+C.解:令lnx=t,则x=e^t∫cos(lnx)dx=∫costd(e^t)=e^t*cost-∫e^tdcost=e^t*cost+∫e^t*sintdt=e^t*cost+∫sintd(e^t)=e^t*cost+e.

∫e^x/(1+e^x) dx=∫1/(1+e^x) dex=∫1/(1+e^x) d(e^x+1)=ln(e^x+1)+C C为任意实数扩展资料不定积分是在微积分中,一个函数f的不定积分,或原函数,或反导数,是一个导数.

用分部积分,原式=2xe^(-2x)+∫e^(-2x)dx=-xe^(-2x)-(1/2)e^(-2x)+c.

没可能降幂吧?醒目点的都知道要用积化和差公式啦! cosxcosy=(1/2)[cos(x+y)+cos(x-y)] ∫[cosx*cos(x/2)]dx =1/2*∫[cos(x+x/2)+cos(x-x/2)]dx =1/2*∫[cos(3x/2)+cos(x/2)]dx =1/2*∫cos(3x/2)dx+1/2*∫cos(x/2)dx d(3x/2)=(3/2)dx,d(x/2)=(1/2)dx =1/2*2/3*∫cos(3x/2)d(3x/2)+1/2*2*∫cos(x/2)d(x/2) =1/3*sin(3x/2)+sin(x/2)+C =(1/3)sin(3x/2)+sin(x/2)+C

∫ (tan2x + sec2x)² dx= ∫ (tan²2x + 2sec2xtan2x + sec²2x) dx= (1/2)∫ (sec²2x - 1 + 2sec2xta2x + sec²2x) d(2x)= (1/2)(2tan2x - 2x + 2sec2x) + c= tan2x + sec2x - x + c

tan^10x*sec^2x = tan^10x d tan x 不定积分 tan^10x*sec^2x dx = (1/11) (tanx)^11

这是公式 ∫ (sec^2x - csc^2x) dx= ∫ sec^2x dx - ∫ csc^2x dx= tanx + cotx + c 欢迎采纳,不要点错答案哦╮(╯◇╰)╭

∫[tanx/(secx)^2dx=∫(sinx/cosx)(cosx)^2dx=∫sinxcosx=∫sinxd(sinx)=(1/2)(sinx)^2+C.