(cosx)^2 (1-(sinx)^3)的积分

cos-x自己画个坐标就一目了然了

2sin²x+cosx-1≥02(1-cos²x)+cosx-1≥02-2cos²x+cosx-1≥0-2cos²x+cosx+1≥02cos²x-c

(sin(cosx^2))'=cos(cosx^2)*(cosx^2)'=cos(cosx^2)*(-sinx^2)*2x=-[2xcos(cosx^2)*sinx^2]

证：(1)(cosx-1)²+sin²x=cos²x-2cosx+1+sin²x=(cos²x+sin²x)+1-2cosx=2-2cosx

tan²-1=sin²x/cos²x-1=(sin²x-cos²x)/cos²x=(sinx+cosx)(sinx-cosx)/cos&su

(1-cosx)'=sinx[(1-cosx)^2]'=(1-cosx)'*2(1-cosx)=2sinx(1-cosx)[sin(1-cosx)^2]'=[(1-cosx)^2]'*cos(1-co

左边=sin²x/(sinx-cosx)-(sinx+cosx)/(sin²x/cos²x-1)=sin²x(sinx+cosx)/(sinx-cosx)(si

∫sin(2x)/(1+cosx)dx=∫2sinxcosxdx/(1+cosx)=-2∫cosxd(cosx)/(1+cosx)=-2∫cosxd[ln(1+cosx)]使用分部积分法得到下一步=-

已知sinx+cosx=1/5sin^2x+2sinxcosx+cos^2x=1/252sinxcosx=-24/25sin^2x-2sinxcosx+cos^2x=1/25-4sinxcosx(si

1.y=√2sin(x+π/4),所以值域是[-√2,√2]2.y=1-cos^2x-cosx+1=-cos^2x-cosx+2设t=cosx,y=-t^2-t+2(=-1

=√{[sin(x/2)]^2+[cos(x/2)]^2-2sin(x/2)cos(x/2)}+√{[sin(x/2)]^2+[cos(x/2)]^2+2sin(x/2)cos(x/2)}=√{[si

sinx|sinx|+cosx|cosx|=-1因为sin²x+cos²x=1所以-sin²x-cos²x=-1所以两个绝对值内都不大于0sinx

∫(0->π/2)(1+cosx)²sin³x(1+2cosx)dx=∫(0->π/2)(1+2cosx+cos^2(x))sin³x(1+2cosx)dx=∫(0->π

cosx-1=1-cos^2xcos^2x+cosx-2=0(cosx+2)(cosx-1)=0因cosx+2>0所以cosx-1=0cosx=1x=2kπ,k为整数.集合就自己表示吧.

原式通分=[(sinx-cosx)²+(sinx+cosx)²]/(cosx+sinx)(cosx-sinx)=2(sin²x+cos²x)/(cos²

sin^2x/(sinx-cosx)-(sinx+cosx)/(tan^2x-1)=sin^2x/(sinx-cosx)-(sinx+cosx)/[(tanx+1)(tanx-1)]=sin^2x/(

1-cosx=1-cos(x/2+x/2)=1-cos²x/2+sin²x/2=2sin²x/2

=e^lim(1/sin²x)·lncosx=e^lim(cosx-1)/x²=e^lim-(1/2)x²/x²=e^-(1/2)

sinx+cosx/sinx-cosx=2化简得sinx=3cosxsin²=9cos²x解出sin²=9/10sinxcosx=3/10sin^2x+2sinxcosx

∫(0→π)sin²x(1+cosx)dx=∫(0→π)sin²xdx+∫(0→π)sin²xcosxdx=∫(0→π)(1-cos2x)/2dx+∫(0→π)sin&#