1 (sinx cosx)^2求不定积分

y=cos2x+sin2x=√2(√2/2*sin2x+√2/2cos2x)=√2(sin2xcosπ/4+cos2xsinπ/4)=√2sin(2x+π/4)所以T=2π/2=π值域是[-√2,√2

∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=∫1/(1+sin^4x)d(1/2*sin²x)=(1/2)∫d(sin²x)/[1

有sin(x-45°)=√2／4=sinxcos45°-cosxsin45°,得sinx-cosx=0.5,两边平方得1-2sinxcosx=0.25.sinxcosx=3／8.tanx+1／tanx

1.因为tanx>0,所以sinx和cosx同号,不妨令二者同正sinx/cosx=3,(sinx)^2+(cosx)^2=1联立,得sinx=3√10/10,cosx=√10/10,所以2sinxc

先化简.f(x)=（sin^4x+cos^4x+sin^2xcos^2x）/（2-2sinxcosx）-1/2sinxcosx+1/4cos^2x=【（sin²x+cos²x）&s

f(x)=sinxcosx+(cosx)^2=sin(2x)/2+[1+cos(2x)]/2=(1/2)[sin(2x)+cos(2x)]+1/2=(√2/2)sin(2x+π/4)+1/2最小正周期

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=

把tanx换成sinx/cosx(cosx+1)(1-cosx)=2sinxcosx1-cos^x=2sinxcosx所以答案是1呦~

x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3

sinxcosx=(1/2)*sin（2x）=(1/2)*{2tanx/[1+(tanx)^2]}=2/5

（1）∵tanx=2，∴原式=1+tanx1−tanx=1+21−2=-3；（2）将已知等式两边平方得：（sinx+cosx）2=1+2sinxcosx=49，即sinxcosx=-518，则sin4

sin^2x+2sinxcosx-3cos^2x=12sinxcosx-4cos^2x=0sinxcosx=2cos^2xcosx（sinx-2cosx）=0（1）cosx=0,sinx=±1（2）s

题目有点怪!会不会打错了啊?提示你哦：sin^2xcos^2x=sin^2(2x)/42sinxcosx=sin2x1/2sinxcosx=sin2x/41/4cos2x=1/4-sin^2x/2再问

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∫dx/(sinxcosx)=∫dx/(tanx*cosx^2)=∫dtanx/tanx=ln|tanx|+C∫dx/(sinxcosx)=∫d2x/(sin2x)=∫csc2xd2x=ln|csc2

sin2x=(2tanx)/(1+tan^2x)=4/5(1)sinxcosx=1/2sin2x=2/5(2)(2sin^2x-3sinxcosx-4cos^2x)/(sinxcosx)=2tanx-

首先把分母化为1=(sinx)^2+(cosx)^2所以原式就为[(sinx)^2+(cosx)^2]/[2sinxcosx+(cosx)^2],然后分子分母同除以一个(cosx)^2式子就可以化为[

hiy=1/2cos^2x+sinxcosx+3/2sin^2x=1/2*（（1+cos2x）/2）+1/2*sin2x+3/2*((1+cos2x)/2)=1/2*sin2x-1/2*cos2x+1

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