17.-|||-设 (a,b)=(int )_(0)^2pi ((acos x-2bsin x))^2dx, 在 (a,b)leqslant 4pi 下,求使得 ^2+4(b)^2-2a-bleqslant -|||-k成立的k的最小值.

(int )_(0)^pi ((xsin x))^2dx; ;

设 M=int_(-(pi)/(2))^(pi)/(2) (sin x)/(1+x^2) cos^4 x dx，N=int_(-(pi)/(2))^(pi)/(

[题目]计算定积分: (int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((|x|+sin x))^2dx

(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}dx= )(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}d

(int )_(0)^2pi |sin x|dx.

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

设= (x,y)|{x)^2+(y)^2leqslant 1} , 则积分 (iint )_(D)((x)^3y+2)dsigma = () .-|||-(A)

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

(B) (int )_({x)_(1)}^(x_{3)}pi [ (f)^2(x)-(g)^2(x)] dx.-|||-(C)(int )_(x)^(x_{3)

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