#int sqrt(1-sqrt(1-x^2)) "d"x#.
Note that
#(1/sqrt(2) (sqrt(1-x)-sqrt(1+x)))^2=1-sqrt(1-x^2)#.
(I found this by setting #(a+bsqrt(1-x^2))^2=1-sqrt(1-x^2)#, expanding the left hand side, equating coefficients and solving for #a# and #b#).
Then,
#sqrt(1-sqrt(1-x^2))=1/sqrt(2) abs(sqrt(1+x)-sqrt(1-x))#.
#int sqrt(1-sqrt(1-x^2)) "d"x= 1/sqrt(2) int abs(sqrt(1+x)-sqrt(1-x)) "d"x#.
The function #sqrt(1+x)-sqrt(1-x)# is real only for #-1<=x<=1#.
Clearly #1+x>1-x# for #0<x<=1#. As the square root function is increasing, #sqrt(1+x)>sqrt(1-x)# or #sqrt(1+x)-sqrt(1-x)>0# for #0<x<=1#.
#1-x>1+x# for #-1<=x<0#. As the square root function is increasing, #sqrt(1-x)>sqrt(1+x)# or #sqrt(1-x)-sqrt(1+x)>0# for #-1<=x<0#.
We conclude,
#sqrt(1-sqrt(1-x^2)) = { (1/sqrt(2)(sqrt(1+x)-sqrt(1-x)), 0<=x<=1), (1/sqrt(2)(sqrt(1-x)-sqrt(1+x)), -1<=x<=0) :}#,
#int sqrt(1-sqrt(1-x^2)) "d"x = { (sqrt(2)/3((1+x)^(3/2)+(1-x)^(3/2))+C, 0<x<=1), (-sqrt(2)/3((1-x)^(3/2)+(1+x)^(3/2))+C, -1<=x<0) :}#
Note that the result is the same but positive for #x>0# and negative for #x<0#. We can define the function #"sign"(x)# to return #+1# for #x>0#, #-1# for #x<0# and be undefined at #0#. We then conclude,
#int sqrt(1-sqrt(1-x^2)) "d"x = "sign"(x) sqrt(2)/3 ((1+x)^(3/2)+(1-x)^(3/2))+C#
If you don't like this #"sign"(x)# function you can alternatively define it as #sqrt(x^2)/x# and conclude,
#int sqrt(1-sqrt(1-x^2)) "d"x = sqrt(2x^2)/(3x)((1+x)sqrt(1+x)+(1-x)sqrt(1-x))+C#
