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∫/(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...

答案给你: ∫1/sinx dx+cosx =∫1/[2sin(x/2)cos(x/2)] dx+sinx =∫1/[sin(x/2)cos(x/2)] d(x/2)+sinx =∫1/tan(x/2)*sec²(x/2) d(x/2)+sinx =∫1/tan(x/2) d[tan(x/2)]+sinx =ln|tan(x/2)|+sinx+C 积分发展的动力来自于实际应用中的需求。实际...

直接凑微分: sinx dx = - d(cosx) = - d(1 + cosx) 所以 ∫ sinx/(1 + cosx) dx = - ∫ d(1 + cosx)/(1 + cosx) = - ln| 1 + cosx | + C.

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

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!

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...

∫1/cosxdx =∫ cosx/cos²xdx =∫ 1/(1-sin²x) d(sinx) =(1/2)∫ [1/(1+sinx)+1/(1-sinx)] d(sinx) =(1/2) [ln(1+sinx)-ln(1-sinx)] + C =ln √[(1+sinx)/(1-sinx)] + C =ln √(1+sinx)²/√(1-sin²x) + C =ln |(1+sinx)/cosx| + C ...

∫ sinx/(1+sinx+cosx) dx =∫sinx(sinx+cosx-1)/[(sinx+cosx+1)(sinx+cosx-1)] dx =∫(sin^2x+sinxcosx-sinx)/[(sinx+cosx)^2-1] dx =∫(sin^2x+sinxcosx-sinx)/(2sinxcosx) dx =(1/2)∫sinx/cosx dx+(1/2)∫ dx-(1/2)∫1/cosx dx =(-1/2)∫1/cosx d(co...

∫sinx/(1+cosx)dx =-∫1/(1+cosx)d(cosx+1) =-ln(cosx+1)+C

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