关于函数=cos x-dfrac (1)(2)cos 2x的极值结论错误的是( ).A 极大值=cos x-dfrac (1)(2)cos 2xB 极小值=cos x-dfrac (1)(2)cos 2xC 极大值=cos x-dfrac (1)(2)cos 2xD 极小值=cos x-dfrac (1)(2)cos 2x

(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}

.int dfrac (1+{cos )^2x}(1+cos 2x)dx

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx

cos dfrac (x)({2)^n})（提示：lim _(xarrow infty )(cos dfrac (x)(2)cos dfrac (x)(4)co

已知曲线积分(int )_(1)^1(1+dfrac ({cos )^2}({x)^2}cos dfrac (y)(x))dx+(sin dfrac (y)(x

(int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx=( ) .(int )_(0)^dfrac (pi {4)}dfr

int dfrac (cos 2x)(cos x+sin x)dx

[题目]如果 (cos x)=dfrac ({sin )^2x}(cos 2x), 则 f(x)= () )-|||-

极限lim _(xarrow 0)dfrac (1-sqrt {1-{x)^2}cos x}(1+{x)^2-(cos )^2x}=极限

[题目] lim _(xarrow 0)(dfrac (1)({sin )^2x}-dfrac ({cos )^2x}({x)^2})= __