5.12 求下列各积分之值:-|||-(1) (int )_(0)^2pi dfrac (dtheta )(a+cos theta )(agt 1);-|||-(2) (int )_(0)^2pi dfrac (dtheta )(5+3cos theta )-|||-(3)(int )_(-x)^ndfrac ({x)^2}({({x)^2+(a)^2)}^2}dx(agt 0);(4)(int )_(-infty )^+infty dfrac (cos x)({x)^2+4x+5}dx;

9.计算下列积分:-|||-(1) (int )_(0)^2pi dfrac (1)(5+3sin theta )dtheta ;-|||-(2) (int

(int )_(dfrac {pi )(4)}^dfrac (pi {2)}dtheta (int )_(0)^cot theta (r)^2cos theta

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}4(cos )^4theta dtheta =__________________

7.计算下列定积分:-|||-(1) (int )_(0)^1x(e)^-xdx ;-|||-(2)(int )_(1)^xxln xdx;-|||-(3) (

πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T

( 10 ) (int )_(0)^dfrac (pi {4)}(tan )^2theta dtheta ;

(单选题) (int )_(0)^2pi (x)^2cos xdx= ().-|||-

5.证明 (z)=cos (z+dfrac (1)(z)) 用z的幂表出的洛朗展开式中的系数为-|||-_(n)=dfrac (1)(2pi )(int )_(

把复数=((cos dfrac {2pi )(9)+isin dfrac (2pi )(9))}^3转化为三角形式A =((cos dfrac {2pi )(9

定积分(int )_(0)^dfrac (pi {2)}(e)^xsin xdx的值是(int )_(0)^dfrac (pi {2)}(e)^xsin xdx