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

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

则 int xf(x)dx= () .-|||-A、 dfrac (1-ln x)({x)^2}+C B、 dfrac (1)(x)+c C、 ln x-x+C

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

若 (x)=xf(xi ), 则 lim _(xarrow 0)dfrac ({s)^2}({x)^2}=-|||-A)1. (B) dfrac (2)(3).

设f(x)为连续函数，则(int )_(0)^1f(dfrac (x)(2))dx等于( )．(int )_(0)^1f(dfrac (x)(2))dx设f(x

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

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

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}((cos )^2x+dfrac (xcos x)(1+{cos )^2x})dx

(int )_(-dfrac {pi )(2)}^dfrac (pi {2)}dfrac (|x|sin x)(1+{cos )^3x}dx=(int )_(-

16、设 (int )_(0)^xf(t)dt=dfrac (1)(2)f(x)-dfrac (1)(2), 其中f(x)为连续函数,则 f(x)=()-|||