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

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

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

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

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

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

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

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

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

计算下列不定积分int dfrac (1)(1+cos 2x)dx计算下列不定积分

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