(8)函数 =dfrac ({x)^2}(sqrt {1-{x)^2}} 在区间 [ dfrac (1)(2),dfrac (sqrt {3)}(2)] 上的平均值为_ __

=dfrac (x)(sqrt {1-{x)^2}}，则=dfrac (x)(sqrt {1-{x)^2}}=_________.，则=_________.

设=dfrac (arcsin x)(sqrt {1-{x)^2}}（1）证明：=dfrac (arcsin x)(sqrt {1-{x)^2}}（2）求=df

=dfrac (arcsin x)(x)+dfrac (1)(2)ln dfrac (1-sqrt {1-{x)^2}}(1+sqrt {1-{x)^2}}

求下列函数的自然定义域：(1)y=sqrt(3x+2)；(2)y=dfrac(1)(1-{x)^2}；(3)y=dfrac(1)(x)-sqrt(1-(x)^2

(6) () =dfrac (1)(sqrt {1-{x)^2}} int dfrac (1)(sqrt {1-{x)^2}}dx=() .

[题目]求下列函数的自然定义域:-|||-(1) =sqrt (3x+2);-|||-(2) =dfrac (1)(1-{x)^2}-|||-(3) =dfra

1.求下列函数的自然定义域:-|||-(1) =sqrt (3x+2);-|||-(2) =dfrac (1)(1-{x)^2};-|||-(3) =dfrac

1.求下列函数的自然定义域:-|||-(1) =sqrt (3x+2) ;-|||-(2) =dfrac (1)(1-{x)^2}-|||-(3) =dfrac

[题目]-|||-int (dfrac (3)(1+{x)^2}-dfrac (2)(sqrt {1-{x)^2}})dx

函数 =arcsin sqrt (1-{x)^2}+dfrac (1)(sqrt {1-{x)^2}} 的定义域为 __ 。