prove for integrals of arcsin inverse trigonometry function

Question

I had difficulty understanding the formula of integrals of arcsin inverse trigonometry function

I did some research and played around the formula, and finally seem to have proved it. I took a screenshot of my work, just in case anyone else is having trouble understanding the formula and require some prove

Answer 1

HINT

Recall that

$$(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}$$

then by chain rule

$$(\arcsin f(x))'=\frac{f'(x)}{\sqrt{1-(f(x))^2}}$$

with $f(x)=\frac x a$.