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1/CostDt求不定积分

有好几种方法的:最常用的是∫ secx dx = ln|secx + tanx| + C 第一种最快: ∫ secx dx = ∫ secx • (secx + tanx)/(secx + tanx) dx = ∫ (secxtanx + sec²x)/(secx + tanx) dx = ∫ d(secx + tanx)/(secx + tanx) = ln|secx + tanx| + ...

令x=2sint,t∈[-π/2,π/2] 则 √(4-x²)=√(4-4sin²t)=2cost,dx=2costdt ∫1/[x√(4-x²)] dx = ∫2cost/(4sintcost) dt =(1/2)∫csctdt =(1/2)ln|csct-cott|+C =(1/2)ln|[2-√(4-x²)]/x|+C C为任意常数

令x=sint,t∈[-π/2,π/2] 则 √(1-x²)=√(1-1sin²t)=cost,dx=costdt ∫1/[x√(1-x²)] dx = ∫cost/(sintcost) dt =∫csctdt =ln|csct-cott|+C =ln|[2-√(1-x²)]/x|+C C为任意常数

令x=3sect,则dx=3sect*tantdt 原式=∫3sect*tant/(9(sect)^2·3tant)dt =1/9·∫costdt =1/9sint+c 反带回x =1/9*√(1-9/x^2)

令 x = tant, 则 I = ∫xe^arctanxdx/(1+x^2)^(3/2) = ∫tante^t(sect)^2dt/(sect)^3 = ∫sinte^tdt = ∫sintde^t = e^tsint - ∫e^tcostdt = e^tsint - ∫costde^t = e^tsint - e^tcost -∫e^tsintdt = e^t(sint-cost) - I, 解得 I = (1/2)e^t(sint-c...

x=tant,dx=sec²tdt ∫dx/[(贰x^贰+依)(x^贰+依)^(依/贰) ] =∫sec²tdt/[(贰tan²t+依)sect] =∫dt/[cost((贰sin²t/cos²t)+依)] =∫costdt/[((贰sin²t+cost²)] =∫[依/(依+sin²t)]d(sint) =arctan(sint)+C 三...

先分母有理化: ∫1/(1+√(1-x^2)dx=∫(1-√(1-x^2)/x^2dx=∫dx/x^2-∫√(1-x^2)/x^2dx 对第二个积分,由分部积分: ∫√(1-x^2)/x^2dx=∫√(1-x^2)d(1/x)=√(1-x^2)/x-∫(1/x)d√(1-x^2)=√(1-x^2)/xarcsinx+C 所以:∫1/(1+√(1-x^2)dx=-1/x+√(1-x^2)/x+arcsinx+C

∫√(1-x²)/x²dx 令x=sint,则√(1-x²)=cost, dx=cost 原式=∫cost/(sint)^2*costdt =∫(cott)^2dt =∫[(csct)^2-1]dt =cott-t+c 带回x, cott=√((1/sint)^1-1)=√(1-x^2)/x, t=arcsinx =√(1-x^2)/x-arcsinx+c 定积分带入积分区间即可

参考一下

令x=sint -π\2

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