Hello,
Answer.
#int arcsin(x)/sqrt(1+x) dx = 2 arcsin(x) sqrt(1+x) + 4 sqrt(1-x) + c#,
where #c in RR#.
Use integration by parts formula :
#int u' v = uv - int u v'#
You can prove that if you write #(uv)' = u'v + uv'#, and integrate that.
Here, #u(x)= arcsin(x)# and #v(x)=2 sqrt(1+x)#. You have :
#u'(x) = 1/sqrt(1-x^2)# and #v'(x) = 1/sqrt(1+x)#.
Apply the formula :
#int arcsin(x) \cdot 1/sqrt(1+x) dx = 2 arcsin(x) sqrt(1+x) - 2 int ( sqrt(1+x))/(sqrt(1-x^2)) dx#
Simplify # ( sqrt(1+x))/(sqrt(1-x^2)) = ( sqrt(1+x))/(sqrt((1-x)(1+x)))= 1/sqrt(1-x)#.
So, you can write
#int arcsin(x) \cdot 1/sqrt(1+x) dx = 2 arcsin(x) sqrt(1+x) - 2 int 1 /sqrt(1-x) dx#
Finally, you know that #int 1/sqrt(1-x) dx = int (1-x)^{-1/2} dx = - 2(1-x)^(1/2) +c# and the result is proved !
