Hello,
Answer.
int arcsin(x)/sqrt(1+x) dx = 2 arcsin(x) sqrt(1+x) + 4 sqrt(1-x) + c∫arcsin(x)√1+xdx=2arcsin(x)√1+x+4√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 !
