∫cos2x (1 sinxcosx) dx

cos(x+π/6)=cosxcosπ/6-sinxsinπ/6所以y=sinx(√3/2*cosx-1/2*sinx)-1/2*cos2x=√3/2*sinxcosx-1/2sin²x-1

令sinxdx=-d(cosx)t^3/(1+t^2)dt=[(t^3+t)-t]/(1+t^2)*dt=t-t/(1+t^2)t^2/2-1/2*ln(1+t^2)+Ccosx^2/2-ln(1+c

cos²x=1-sin²x=(1+sinx)(1-sinx).∴y=2sinx(1-sinx)=-2(sin²x-sinx)=-2[sinx-(1/2)]²+(

Letu=1+sin(x)cos(x)=1+(1/2)sin(2x)anddu=cos(2x)dx→dx=du/cos(2x)So∫cos(2x)/(1+sin(x)cos(x))dx=∫1/udu=

已知sin2x=2sinxcosxcos2x=(cosx)^2-(sinx)^2所以1-cos2x=2(sinx)＾21+cos2x=2(cosx)^2所以（1+cos2x)/2cosx=sin2x/

原式＝(-2cos2x/1+sin2x+cos2x)+1=(-2cos^2x+2sin^2x)/(1+2sinxcosx+cos^2x-sin^2x)+1=[2(sinx+cosx)(sinx-cos

答：y=sin2x+2sinxcosx+2cos2x=2sin2x+2cos2x=2√2sin(2x+π/4)1）当sin(2x+π/4)=-1时,y的最小值为-2√22x+π/4=2kπ-π/2,x

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1+sin2x-cos2x=1+2sinxcosx-1+2sinx^2=2sinx(cosx+sinx)1+sin2x+cos2x=1+2sinxcosx+2cosx^2-1=2cosx(cosx+s

能不能加上括号?y=(2sinxcos^2x)/(1+sinx)=[2sinx(1-sin^2x)]/(1+sinx)=[2sinx(1+sinx)(1-sinx)]/(1+sinx)=2sinx(1

y=2sinXcos^2X/1+sinX=2sinx(1-sin^2x)/(1+sinx)=2sinx(1+sinx)(1-sinx)/(1+sinx)=2sinx-2sin^2x令sinx=t,-1

左式中1=(sinx平方+cosx平方-2sinxcosx)/(cosx+sinx)(cosx-sinx),\x0d约去cosx-sinx后,\x0d=(cosx-sinx)(cosx+sinx),然

1+cos2x=2cos²分母为cos²2x*cos²x*(1+cosx)分子为4sinx*cos²2x*cos²x化简为4sinx/(1+cosx)

∫(1+（cos）^2x)/(1+cos2x)dx=∫(1+（cos）^2x)/（2cos^2x）dx=∫[1/(2cos^2x)+1/2]dx=x/2+∫1/(2cos^2x)=(x+tanx)/2

y=2sinxcos^2x/1-sinx=2sinx(1+sinx)=2(sinx+1/2)^2-1/2-1/2≤sinx+1/2≤3/20≤(sinx+1/2)^2≤9/4函数y=2sinxcos^

∫（COS2X）／（1十SinXCOSX）dX=∫(1/2)/(1+sin2x/2)d(sin2x)=∫(1/2)/(1+u/2)du(u=sin2x)=∫1/(u+2)d(u+2)=ln|u+2|+

(cos2x-sin2x)/[(1-cos2x)(1-tan2x)]=cos2x[1-(sin2x/cos2x)]/[(1-cos2x)(1-tan2x)]（分母部分提出cos2x）=cos2x(1-

cos2x=2cos^2x-1所以1+cos2x=2cos²x再问：。。。。为什么∫(1+cos²x)/(2cos²x)dx=x/2+(1/2)∫dx/cos²