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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

(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)]/(...

积分部分 =2/[(cosx/sinxcosx)+(sin^2x/cosxsinx)] =2/[(1/sinx)+(sinx/cosx] =2/(cscx+tanx)

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...

=∫(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

请看图

1、本题运用变量代换 substitution,也就是换元法,解答如下图; 2、楼主若有疑问,欢迎追问,有问必答,有疑必释; 3、若点击放大,图片将会更加清晰。 . . 【敬请】敬请有推选认证《专业解答》权限的达人,千万不要将本人对该题的解答认证为《...

∫(cos⁵x·sin2x)dx =2∫(cos⁵x·sinxcosx)dx =2∫(cos⁶x)(sinxdx) =-2cos⁶xd(cosx) =(-2/7)cos⁷x+C ∴∫(0,π/2)(cos⁵x·sin2x)dx =(-2/7)cos⁷(π/2)-(-2/7)cos⁷(0)=2/7

1.设u=tan(1/2*x),所以sinx=2u/(1+u^2),cosx=(1-u^2)/(1+u^2),dx=2/(1+u^2)du 代入化简得: 原式=∫1/u(3-u^2)du =∫1/3*1/udu+∫2/3*1/(3-u^2)du =-2/3*lnlu^2-3l+1/3*lnlul+C =-2/3*lnltan(1/2*x)^2-3l+1/3*lnltan(1/2*x)l+C 2.分子分母同乘sinx, ...

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