∫(cos2x)^3sin(2x)dx

f(x)＝2sin²[(π/4)－x]－√3cos2x＝1－cos[(π/2)－2x]－√3cos2x＝1－sin2x－√3cos2x＝－2sin(2x＋π/3)＋1∵－1≤sin(2x＋π

f(x)＝2sin2xcosπ3+cos2x＝sin2x+cos2x=2sin(2x+π4)…（4分）（1）函数f（x）的最小正周期为2π2＝π…（6分）（2）由题意知g（x）=f（x+3π8）+2=

函数f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)1.当x属于R时,函数f(x)的最小正周期T

5.（1）f(x)=(√3)cos2x+2sinxsin(x+π/2).=√3cos2x+2sinxcosx=sin2x+√3cis2x=2sin(2x+π/3).∴最小正周期T=2π/2=π,f(x

函数f(x)=[2sin^2(x+π/4)-1]-√3cos2x+1=[-cos(2x+π/2)]-√3cos2x+1=sin2x-√3cos2x+1=2(1/2*sin2x-√3/2*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/(sin²x*cos²x)dx=∫cos2x/(1/2*sin2x)²dx=4∫cos2x/(sin²2x)dx=4∫csc2x*cot2xdx=

/>f(x)=sin(2x+3π/2)=sin(2x+3π/2-2π)=sin(2x-π/2)=-sin(π/2-2x)=-cos2x

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

cos2x+cos(x+π/3)+sin(x+π/6)+3sin^2x=cos2x+1/2*cosx-根号3/2*sina+根号3/2*sinx+1/2*cosx+3sin^2x=cos2x+cosx

解题思路：利用差角正弦展开，合并，化为正弦型函数，利用周期公式。。。。。。。。。。。。解题过程：

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

f(x)=2sin^2((π/4)-x)-(√3)(cos2x)=1-cos(π/2-2x)-(√3)cos2x=1-sin2x-(√3）cos2x=1-2sin(2x+π/3),(1)f(x)的最小

f(x)=√3sin2x＋cos2x＋2f(x)=2sin(2x＋π/6)＋21、函数f(x)的最大值是4,此时：2x＋π/6=2kπ＋π/2即：x=kπ＋π/6自变量x的取值集合是：{x|x=kπ＋

y=sin(2x-π/3)+根号3cos2x=sin2xcosπ/3-cos2xsinπ/3+2sinπ/3cos2x=sin2xcosπ/3+cos2xsinπ/3=sin(2x+π/3)1.振幅=

f(x)=2sin^2(pi/4+x)-√3cos2x=[1-cos(2(pi/4+x)]-根号3cos2x=1-cos(pi/2+2x)-根号3cos2x=1+sin2x-根号3cos2x=1+2[

∵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．

sin(2x+3π/2)=sin2xcos3π/2+cos2xsin3π/2cos3π/2=0,sin3π/2=-1原式=-cos2x