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

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

( (int )_(dfrac {pi )(4)}^dfrac (pi {3)}dfrac (x)({sin )^2x}dx ;

(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 )_(-

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

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

7.计算下列定积分.-|||-(3) (int )_(dfrac {pi )(4)}^dfrac (pi {3)}dfrac (x)({sin )^2x}dx

例: (int )_(0)^dfrac (pi {2)}(cos )^2dfrac (x)(2)dx= __ .

int dfrac (1)(1+cos 2x)dx=______.______.

( 20 int dfrac (dx)(1+cos 2x) ;