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∫(sinx+sin2x)/(1+Cosx的平方)

∫ (x + 2sinxcosx)/(1 + cos2x) dx = ∫ x/(1 + cos2x) dx + ∫ 2sinxcosx/(1 + cos2x) dx = ∫ x/(1 + 2cos²x - 1) dx + ∫ sin2x/(1 + cos2x) dx = (1/2)∫ xsec²x dx - (1/2)∫ d(cos2x)/(1 + cos2x) = (1/2)∫ x d(tanx) - (1/2)∫ d(1 + ...

∫sinx/(1+sin^2x)dx=-∫d(cosx)/[2-(cosx)^2] =-∫d(cosx)/(√2+cosx)(√2-cosx) =∫d(cosx)/(√2+cosx)(cosx-√2) =√2/4∫[1/(cosx-√2)-1/(cosx+√2)]dcosx =√2/4ln|(cosx-√2)/(cosx+√2)|+C

左边=cos²x+2sinxcosx+sin²x=1+2sinxcosx=1+sin2x=右边。

变形=2sinxcosx/[cosx+(1-cos2x)/2]dx =-2cosx[cosx+(1-cos2x)/2]d(cosx) 令t=cosx

(sin2x/sinx-cosx)-(sinx+cosx/tan2x-1) =[sin2x*(sinx+cosx)]/[(sinx-cosx)(sinx+cosx)]-(sinx+cosx)/[(sin2x/cos2x)-1] =[sin2x*(sinx+cosx)]/(sin2x-cos2x)-[cos2x(sinx+cosx)]/(sin2x-cos2x) =[(sinx+cosx)(sin2x-cos2x)]/(...

=∫(1-2sin²x-2sinxcosx)/(cosx+sinx)dx =∫1/√2sin(x+π/4)-2sinxdx =-1/√2∫1/(1-cos²(x+π/4))dcos(x+π/4)+2cosx =-(1/2√2)ln(1+cos(x+π/4))/(1-cos(x+π/4))+2cosx+C

lim(x->0)(1+sinx-cosx)/(1+sin2x-cos2x) (0/0) =lim(x->0)(cosx+sinx)/(2cos2x+2sin2x) =1/2

解: 1/sinx+1/cosx=2√2 (sinx+cosx)/(sinxcosx)=2√2 (sinx+cosx)²/(sinxcosx)²=(2√2)² (1+sin2x)=(√2sin2x)² 令sin2x=t, 则,1+t=2t² 2t²-t-1=0 (2t+1)(t-1)=0 所以, t=-1/2或t=1 所以, cos2x=±√3/2或0 sin(2...

sin2x+2sinxcosx+cos2x可以等于1 我们可以证明: sin2x+2sinxcosx+cos2x =(sinx+cosx)2 =【根号2sin(x+π/4)】2 =2sin2(x+π/4) 当x=kπ的时候2sin2(x+π/4)=1成立 这与sin2x+cos2x并不矛盾 因为这和x的取值有关系

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