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∫(1/Cos²x)Dx怎么求解

(tanx)'=1/cos²x ∴∫(1/cos²x)dx=tanx+C

∫secx^3dx =∫secxdtanx =secxtanx-∫tanxdsecx =secxtanx-∫(secx^2-1)secxdx =secxtanx+∫secxdx-∫secx^3dx 2∫secx^3dx=secxtanx+∫secxdx=secxtanx+ln|secx+tanx| ∫secx^3dx=(1/2)secxtanx+(1/2)ln|secx+tanx|+C

这是基本公式,等于tanx+c

cos2x=2cos^2x-1 所以 1+cos2x=2cos²x

直接用基本积分公式。经济数学团队帮你解答。请及时评价。谢谢!

∫ x/(1+cosx) dx =∫ x/[2cos²(x/2)] dx =∫ xsec²(x/2) d(x/2) =∫ x dtan(x/2) 分部积分 =xtan(x/2) - ∫ tan(x/2) dx =xtan(x/2) - 2∫ sin(x/2)/cos(x/2) d(x/2) =xtan(x/2) + 2∫ 1/cos(x/2) dcos(x/2) =xtan(x/2) + 2ln|cos(x/2)| + C...

#include #include #define PI (acos(-1))#define STEP (1e-6)double func(double x);double inte(double up,double down,double func(double));int main(void) {double up,down;printf("%lf%lf",&up,&down);printf("%lf\n",inte(1,0,func));ret...

把log改成ln就可以了

吧cosx看做t 则原式=∫dt/(1+他) 线条他等于arctant+C 吧t换成cosx 所以原式=arctan(cosx)+C

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