求四道关于三角函数的不定积分 三角函数求不定积分,求助!
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∫ dx/sinx
=∫ cscx dx
=ln|cscx-cotx| + C
(2)
∫ dx/cosx
=∫ secx dx
=ln|secx+tanx| +C
(3)
∫ dx/(sinx)^2
=∫ (csx)^2 dx
=-cotx + C
(4)
∫ dx/(sinx)^3
=∫ (cscx)^3 dx
=-∫ cscx d(cotx)
=-cscx.cotx - ∫ (cotx)^2.cscx dx
=-cscx.cotx - ∫ [(cscx)^2 -1].cscx dx
2∫ (cscx)^3 dx =-cscx.cotx + ∫ cscx dx
∫ (cscx)^3 dx =(1/2)[-cscx.cotx + ln|cscx-cotx| ] + C
=>
∫ dx/(sinx)^3 =(1/2)[-cscx.cotx + ln|cscx-cotx| ] + C
前3个有公式
∫dx/sinx = ∫cscxdx = ln|cscx - cotx| + C
∫dx/cosx = ∫secxdx = ln|secx + tanx| + C
∫dx/(sinx)^2 = ∫(cscx)^2dx = -cotx + C
∫dx/(sinx)^3 = -∫sinxdx/(sinx)^4
= -∫dcosx/[1-(cosx)^2]^2 记 u = cosx
= -∫du/(1-u^2)^2
= (-1/4)∫[1/(1-u) + 1/(1+u) + 1(1-u)^2 + 1/(1+u)^2]du
= (-1/4)ln|(1+u)/(1-u)| + 1/(1-u) - 1/(1+u) + C
= (-1/4)ln|(1+cosx)/(1-cosx)| + 1/(1-cosx) - 1/(1+cosx) + C
不定积分结果不唯一求导验证应该能够提高凑微分的计算能力,
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绛旓細=1/4ln缁濆鍊紅an(x/2)+1/8{tan(x/2)}^2+C
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绛旓細璁$畻锛坱anx锛²涓嶅畾绉垎鐨勬柟娉曪細锛坱anx锛²=鈭玔(secx)^2-1]dx =鈭(secx)^2dx-x =tanx-x+c
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