. . .-|||-.-dfrac (2)(3)pi +2kpi ,k=0.pm 1 . . . .-|||-(C) dfrac (1)(3)pi +2kpi 一=0,pm 1 . . . .-|||-D 以上都不对

(-3+4i)=ln 5+i(2kpi +pi -arctan dfrac (4)(3))-|||-A.对-|||-B.错 二

) ( )-|||-(A) =pm (2k+1) (B) =pm 2k-|||-(C) =pm dfrac (1)(2)(2k+1) (D) =pm dfrac

dfrac ({mu )_(0)I}(pi R)(dfrac (1)(2)+dfrac (pi )(6))-|||-dfrac ({mu )_(0)I}(pi

3.要使函数φ(x )= ,dfrac {pi )(2)] (B)[π,2π] (C) [ 0,dfrac (pi )(2)] (D) [ dfrac

dfrac (1)(2)pi .-|||-C.π-|||-D. dfrac (5)(4)pi .-|||-0

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

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

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

2.计算下列定积分:-|||-(1) (int )_(dfrac {pi )(3)}^pi sin (x+dfrac (pi )(3))dx;-|||-(2)

1.计算下列定积分:-|||-(1) (int )_(dfrac {pi )(3)}^pi sin (x+dfrac (pi )(3))dx ;-|||-(2)