(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
[题目] (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx= __
[题目] (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx
计算: (int )_(0)^sqrt (2)sqrt (2-{x)^2}dx= .计算: .
(12) int dfrac (x)(sqrt {2)-3(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