(int )_(0)^dfrac (pi {2)}sqrt (1-sin 2x)dx;

(1)求 (int )_(0)^dfrac (pi {2)}sqrt (1-sin 2x)dx;

求定积分 (int )_(0)^dfrac (pi {2)}sqrt (1-sin 2x)dx的值.求定积分 的值.

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

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

求不定积分int dfrac (1)(2x)sqrt (ln x)dx=（）.int dfrac (1)(2x)sqrt (ln x)dx=int dfrac

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

(int )_(1)^2dfrac (dx)(x(1+sqrt {2x))}=( )A.(int )_(1)^2dfrac (dx)(x(1+sqrt {2x

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

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

40) int dfrac (dx)(1+sqrt {2x)} ,