判断：设=sin xcos x，则=sin xcos x.（）。

计算：int dfrac (1)(sin xcos x)dx计算：

求导：y=sin^2xcos2x.求导：$y=sin^2xcos2x$.

[单选题]设函数Y—sin2xcos2x的最小正周期是 （ ）A．AB．BC．CD．D

求下列不定积分：int dfrac (sin xcos x)(1+{sin )^4x}dx.求下列不定积分：.

lim _(xarrow 0)dfrac (sin x-xcos x)({x)^2ln (1+2x)}= ()_________________________

12.求极限lim_(xto0)(x-xcos x)/(tan x-sin x).12.求极限$\lim_{x\to0}\frac{x-x\cos x}{\ta

1.求下列极限.-|||-(1) lim _(xarrow 0)dfrac (x-xcos x)(x-sin x)

【题目】-|||-dfrac (d)(dx)(int )_(sin x)^cos xcos (pi (t)^2)dt .

【题目】-|||-dfrac (d)(dx)(int )_(sin x)^cos xcos (pi (t)^2)dt .

1.[判断题] 判断：设sin y+int_(0)^x^(2)cos tdt=e^y，则(dy)/(dx)=(2xcos x^2)/(cos y-e^y).（）