∫(0→π 2)1 [1 (cosx)∧2]dx

=根号2·∫(0,3π)根号下[(1-cosx)/2]dx=根号2·∫(0,3π)根号下[sin^2(x/2)]dx=根号2·∫(0,3π)|sin(x/2)|dx=2根号2·[∫(0,2π)sin(

∫(0,π/2)[f(cosx)cosx-f'(cosx)sin^2x]dx=∫(0,π/2)d[sinxf(cosx)]=sinxf(cosx)|(0,π/2)=1*f(0)-0*f(1)=f(0)

令u=tan(x/2)=>dx=2du/(1+u²),cosx=(1-u²)/(1+u²)当x=0,u=0//当x=π,u=+∞原式=∫[0,+∞]1/[2+(1-u&#

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[(sinx+cosx-1)(sinx-cosx+1)-2cosx]/sin2x=[(sinx)^2-(cosx-1)^2-2cosx]/sin2x=[(sinx)^2-(cosx)^2-1]/sin

∫(1/x²+1)dx=-1/x+x+C选A

(1+cosx)/(1-cosx)+(1-cosx)/(1+cosx)通分=((1+cosx)^2+(1-cosx)^2)/1-cos^2(x)=2*(1+cos^2(x))/sin^2(x)因为1=

cosx=2cos^2(x/2)-1因为0

y=(sinx/x)^(cosx/1-cosx)lny=(cosx(lnsinx-lnx)/(1-cosx)limlny=lim(cosx(lnsinx-lnx)/(1-cosx)=lim(lnsin

左边通分=(cosx+cos²x-sinx-sin²x)/(1+sinx+cosx+sinxcosx)=[cosx-sinx+(cosx+sinx)(cosx-sinx)]/(si

f(x)=2ab-1=2*(√3sinx,cosx)·(cosx,cosx)-1=2√3sinxcosx+(2cos^2(x)-1)==√3sin2x+cos2x=2sin(2x+π/6)π/6≤2x

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

证明：右边=cosx/(1+sinx)-sinx/(1+cosx)=[cosx(1+cosx)-sinx(1+sinx)]/(1+sinx)(1+cosx)=(cosx-sinx)(1+sinx+co

因为3sinx-2cosx=0,所以sinx/2=cosx/3.令sinx=2k,cosx=3k,k≠0.(1)原式=(3k-2k)/(3k+2k)+(3k+2k)/(3k-2k)=(1/5)+5=2

∫(0→π/4)cos²xdx=∫(0→π/4)(1+cos2x)/2dx=∫(0→π/4)1/2dx+∫(0→π/4)1/2·cos2x(1/2)d(2x)=1/2·π/4+1/4·sin

由原式可得1/(cosx-sinx)^2=4cosx/(cosx-sinx)-2=2(sinx+cosx)/(cosx-sinx)所以1/(cosx-sinx)=2(sinx+cosx)则2(sinx

由[sinx-2cosx][3+2sinx+2cosx]=0可得sinx-2cosx=0或者sinx+cosx=-3/2可因为(sinx+cosx)的最小值为-根号2>-3/2,故sinx+cosx=

∫sinx(cosx+1)dx/[1+(cosx)^2]=-∫(1+cosx)dcosx/[1+(cosx)^2]=-∫dcosx/[1+(cosx)^2]-(1/2)∫d[1+(cosx)^2]/[

F(x)=a*b+1=2cosx^2=2cosx*sinx+1=cos2x+sin2x=√2sin(2x+派/4)剩下自己算即可再问：不好意思，我问的是第二小题。再答：那更不用说了，最小为√2*（-√

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