设 M=int_(-(pi)/(2))^(pi)/(2) (sin x)/(1+x^2) cos^4 x dx,N=int_(-(pi)/(2))^(pi)/(
+dfrac (1)({(2n-1))^2}(2n-1)x+... ] -|||-(B) dfrac (2)(pi )[ dfrac (1)({2)^2}sin
(sin x)=dfrac (1)({cos )^2x} in (0,dfrac (pi )(2)),则(sin x)=dfrac (1)({cos )^2x}
函数(x.y)=4x-(x)^2-(y)^2的极值为( )(x.y)=4x-(x)^2-(y)^2(x.y)=4x-(x)^2-(y)^2(x.y)=4x-(x
一、选择题-|||-3.设 =(sin )^2x, 则 ^(n+1)=() .-|||-(A) sin (2x+dfrac (npi )(2)) (B) ^ns
设=(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (sin x)(1+{x)^2}(cos )^4xdx, =(in
(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx
2.求下列函数在指定点的导数:-|||-(1)设 =f(x)=sin x-cos x, 求 (dfrac (pi )(4)), (dfrac (pi )(2))
设(x)=sin x, 则(dfrac (pi )(2))= __-|||-_.-|||-A cos dfrac (pi )(2)-|||-B o-|||-C-
cos dfrac (x)({2)^n})(提示:lim _(xarrow infty )(cos dfrac (x)(2)cos dfrac (x)(4)co