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∫(Cosx)∧2Dx和∫(Cosx)∧3Dx怎么用积分求?

显然∫(cosx)^2 dx =∫ 1/2 *[2(cosx)^2 -1] +1/2 dx =∫ 1/2 *cos2x +1/2 dx =1/4 *cos4x +x/2 +C 而∫ (cosx)^3 dx =∫ (cosx)^2 d(sinx) =∫1-(sinx)^2 d(sinx) =sinx -1/3 *(sinx)^3 +C

=∫tan²xsecxdx=∫tanxdsecx=tanxsecx-∫secxdtanx=tanxsecx-∫secx(tan²x+1)dx=tanxsecx-∫secxtan²xdx-∫secxdx 所以∫tan²xsecxdx=1/2(tanxsecx-∫secxdx)=1/2(tanxsecx-∫1/(1-sin²x)dsinx)=1/2(tanxsecx-1/2∫1/(1+sinx)+1/...

∫π/2到-π/2√cosx-cosx^3dx =-2∫(0,π/2)√cosx(1-cos²x)dx =-2∫(0,π/2)sinx√cosxdx =2∫(0,π/2)√cosxdcosx =2*(2/3)(cosx)^(3/2)|(0,π/2) =-4/3

∫sinx/(cosx)^3dx =-∫1/(cosx)^3dcosx =-1/(-3+1)(cosx)的(-3+1)次方+c =1/2 (cosx)的-2次方+c =1/2 sec²x+c

给你提供一个思路,3倍角公式cos(3a)=4(cosa)^3-3cosa,还有一个是凑微风法,具体过程我不写了,微风题目一定要多做才能熟悉

设x^3=t 原式=1/3∫tcostdt =1/3∫tdsint =1/3tsint+1/3cost+C =1/3x^3sinx^3+1/3cosx^3+C

终南别业(王维)

原积分=∫[1-3cosx+3(cosx)^2-(cosx)^3]dx =∫(1-3cosx)dx+3∫(cosx)^2dx-∫(cosx)^3dx =x-3sinx+3/2∫(1+cos2x)dx-∫(cosx)^2dsinx =x-3sinx+3/4∫cos2xd2x+3/2∫dx-∫[1-(sinx)^2]dsinx =5/2×x-4sinx+3/4×sin2x+1/3×(sinx)^3+C.

问题一:令X=tgt,则dx=sec²tdt ∫x²(1+x²)^½dx=∫tg²·sect·sec²tdt =∫sin²t/cos^5t·dt =∫(1-cos²t)/cos^5t·dt =∫1/cos^5t·dt-∫1/cos³t·dt =1/(5-1)sint/cos^4t+(5-2)/(5-1)∫1/cos³t·...

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