(14) int dfrac ({x)^2}(sqrt {{a)^2-(x)^2}}dx;

(int )_(0)^sqrt (2)sqrt (2-{x)^2}dx

(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}dx= )(int )_(-pi )^pi sqrt ({pi )^2-(x)^2}d

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

计算: (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx= ．计算: ．

[题目] (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx

[题目] (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx= __

int dfrac (x)(sqrt {2-3{x)^2}}dx。。

int dfrac (x)(sqrt {2-3{x)^2}}dx

(2) int dfrac (sqrt {{x)^2-9}}(x)dx

(12) int dfrac (x)(sqrt {2)-3(x)^2}dx;