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∫1/(1+sinx%Cosx)Dx

采用换元法与分部积分法,及基本的积分公式表 下面是总结积分题的方法:

这个是三角函数的不定积分,分母应先进性化简,计算步骤为: ∫1/(sinx+cosx)dx =∫dx/√2sin(x+π/4) =-(√2/2)∫dcos(x+π/4)/sin^2(x+π/4) =-(√2/4){∫dcos(x+π/4)/[1-cos(x+π/4)]+∫dcos(x+π/4)/[1+cos(x+π/4)]} =-(√2/4)ln{[1+cos(x+π/4)]/[1-cos...

你好!下面提供两种做法,都是第一类换元法。经济数学团队帮你解答,请及时采纳。谢谢!

先求不定积分 ∫1/sinx dx =∫sinx/sin²xdx =-∫1/sin²xdcosx =-∫1/(1-cos²x)dcosx =∫1/(cosx+1)(cosx-1)dcosx =∫[1/(cosx-1)-1/(cosx+1)]/2dcosx =[∫1/(cosx-1)dcosx-∫1/(cosx+1)dcosx]/2 =[∫1/(cosx-1)d(cosx-1)-∫1/(cosx+1)d(cos...

这个更好。

∫(1+sinx)/[sinx(1+cosx)]dx =∫1/[sinx(1+cosx)]dx+ ∫1/(1+cosx)dx =∫sinx/[sin^2x(1+cosx)]dx+ ∫(1-cosx)/(1-cos^2x)dx =-∫1/[(1-cos^2x)(1+cosx)]dcosx+ ∫(1/sin^2xdx- ∫1/sin^2xdsinx =-∫1/[(1-cosx)(1+cosx)^2]dcosx-ctnx+1/sinx =-∫1/[(1-...

用万能代替 ∫1/(sinx+cosx)dx =∫1/{2tan(x/2)/[1+tan^2(x/2)]+[1-tan^2(x/2)]/[1+tan^2(x/2)]}dx =∫[1+tan^2(x/2)]/[2tan(x/2)+1-tan^2(x/2)]dx =-∫1/[-2tan(x/2)-1+tan^2(x/2)]dtan(x/2) =-∫1/{[tan(x/2)-1]^2-2}dtan(x/2) =-1/(2√2)∫{1...

原式=∫(1+cosx)dx/[1-(cosx)^2] =∫(1+cosx)dx/(sinx)^2 =∫dx/(sinx)^2+∫cosxdx/(sinx)^2 =-cotx+∫d(sinx)/(sinx)^2 =-cotx-1/sinx+C.

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