∫{［(cos2x) (sin²xcos²x)］}dx-搜答案-百度作业网

①原式=f(x)=2cos2x+sinx^2=2cos2x+1-cos2x/2=3/2cos2x+1/2故f(π/3)=3/2*cos2π/3+1/2=-3/4+1/2=-1/4②依f(x)=3/2c

y=2(1-2sin²x)+sin²x-4cosx=2-3sin²x-4cosx=2+3cos²-3-4cosx=3cos²-4cosx-1;再问：-

诱导公式sin（a+π/2）=cosa所以sin（2x+π/2）=cos2x

再问：好快~而且是图片所以很清楚~赞再答：有点误再问：只是最後答案算错了吗?再答：是的另有简单方法如下：再问：厉害喔~!!谢谢你~🙏再答：做完后发现此题考察是积分函数的绝对值和奇偶性再

cos2x/(cosx+sinx)=(cos²x-sin²x)/(cosx+sinx)=(cosx+sinx)(cosx-sinx)/(cosx+sinx)=cosx-sinx求导

看图片吧

∫cos2x/(sin²x*cos²x)dx=∫cos2x/(1/2*sin2x)²dx=4∫cos2x/(sin²2x)dx=4∫csc2x*cot2xdx=

1.∫cos(2x)/(cos²xsin²x)dx=∫cos(2x)sec²xcsc²xdx=4∫cot(2x)csc(2x)dx=2∫cot(2x)csc(2

1.将分母变为sin2x即原式为∫[（4cos2x/sin^2（2x））]dx2.进行换元即2x变为t,原式变为∫[（2cos2x/sin^2t）]dt.3继续换元,可观察到（sint）'=cost.

2cos2x+sin^2x 化简

2cos2x+sin^2x=2(cos^2x-sin^2x)+sin^2x=2cos^2x-sin^2x=3cos^2x-1

cos2x=sin(π/2-2x)=2sin(π/4-x)cos(π/4-x)cos2x/[sin(π/4-x)]=2sin(π/4-x)cos(π/4-x)/[sin(π/4-x)]=2cos(π/

（Ⅰ）f(x)＝sinx•cosx+12cos2x+12=12sin2x+12cos2x+12=22sin(2x+π4)+12∴函数f（x）的最小正周期T＝2π2＝π（Ⅱ）当x∈[−π8，3π8]时，

即cos^8x-sin^8x=(cos^4x+sin^4x)(cos^4x-sin^4x)=(cos^4x+sin^4x)(cos²x+sin²x)(cos²x-sin&

y=cos2x+sin²x-cosx=cos²x-cosx=(cosx-1/2)²-1/4x=2kπ+π,max(y)=2x=2kπ±π/3,min(y)=-1/4x∈[

∵f（x）=sin（π3−2x）+cos2x=32cos2x-12sin2x+cos2x=（32+1）cos2x-12sin2x=2+3sin（2x+θ）∴T=2π2=π故答案为：π．

f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))=(cos2x)^2/((sinπ/4cosx+cosπ/4sinx)(sinπ/4cosx-cos

因为sin^2(X)+cos^2(X)=1所以原式=1-cos^2（x）-cos^2（x）=1-2cos^2(x)=-(2cos^2(x)-1)=-cos2x

sin(x/2)的周期是4pi,cos2x的周期是pi,sin(x/2)+cos2x的周期是其最小公倍数,自然是4pi

f(sinx-1)=cos2x+2cos2x=1-2sin^2xf(sinx-1)=3-2sin^2x=-2(sinx-3/4)^2+7.5f(x)=-2(x+1/4)^2+7.5