【题目】 (int )_(0)^dfrac (pi {2)}cos xdx= ()-|||-A、 dfrac (1)(2)-|||-B、1-|||-C、 -dfrac (1)(2)-|||-D、 -1

设=(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (sin x)(1+{x)^2}(cos )^4xdx, =(in

[题目]计算定积分: (int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (1)(2)xcos xdx.

(int )_(0)^1dfrac (1)(2)xdx= A.1B.0C.(int )_(0)^1dfrac (1)(2)xdx=D.2A.1B.0C.D.2

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

(int )_(0)^dfrac (pi {4)}(tan )^2xdx=1-dfrac (pi )(4) __-|||-__下列式子或叙述不正确的是A.B.设

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

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (|x|sin x)(1+{cos )^3x}dx=(int )_(-

(3) (int )_(0)^dfrac (pi {4)}dfrac (x)(1+cos 2x)dx= () .-|||-(A) dfrac (pi )(8)+

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

[ dfrac (sin z)({z)^2},0] =-|||-A 1-|||-B .-1-|||-C dfrac (1)(2)