2、设f`(x)为连续函数,则 int (sin )^2xf'(cos x)dx-int cos xf(cos x)dx=

设 f(2)=4 , (int )_(0)^2f(x)dx=1, 则 (int )_(0)^2xf(x)dx=

(1)设 dfrac (cos x)(x) 是f(x)的一个原函数,则 int xf(x)dx= __ ;

计算不定积分int (x+cos x)dx=（）A int (x+cos x)dx=B int (x+cos x)dx= C int (x+cos x)dx=

int dfrac (cos 2x)(cos x+sin x)dx

若 int f(x)dx = sin x + C，则 int xf(1-x^2)dx = ( )A. $2\sin(1-x^2)+ C$B. $-\frac{1

(d)/(dx)int_(sin x)^cos x cos(pi t^2) , dt = ( ). $\frac{d}{dx}\int_{\sin x}^{\c

int dfrac (sin x-cos x)(sin x+cos x)dx==

14.int(sin x+3cos x)/(2sin x+cos x)dx=____14.$\int\frac{\sin x+3\cos x}{2\sin x+

5.设函数 (x)=cos x, 则 (int )_(0)^dfrac (pi {2)}xf(x)dx= ()-|||-【】-|||-A.0 B 1 C -1

)int dfrac (1)({cos )^2x(sin )^2x}dx.