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

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

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

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

5.求不定积分 int dfrac ({sin )^2x}(1+{sin )^2x}dx.

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

求下列定积分:-|||-(int )_(dfrac {sqrt {2)}(2)}^1dfrac (sqrt {1-{x)^2}}({x)^2}dx

( (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

求定积分(int )_(1)^sqrt (3)dfrac (1)({x)^2sqrt (1+{x)^2}}dx求定积分

2.计算下列定积分:-|||-(1) (int )_(dfrac {pi )(3)}^pi sin (x+dfrac (pi )(3))dx;-|||-(2)