Solve the following integral: ∫ arcsin(x)/sqrt(1-x^2) dx

We will solve the integral by part. We know the formula for integration by parts: ∫ f(x)'g(x)dx=f(x)g(x)-∫f(x)g(x)'dx (1). We know that: (arcsin (x))'=1/sqrt(1-x^2). So we can write arcsin(x)/sqrt(1-x^2) dx =arcsin(x)*(arcsin(x))'. So, in formula (1) f(x)=arcsin(x), g(x) =arcsin(x) and f(x)'g(x)=arcsin(x)/sqrt(1-x^2) dx. So, using (1) we obtain: ∫ arcsin(x)/sqrt(1-x^2) dx=∫ (arcsin(x))'*arcsin(x)dx=(arcsin(x))²-∫ arcsin(x)arcsin(x)'dx=(arcsin(x))²- ∫ arcsin(x)/sqrt(1-x^2) dx. We obtained: ∫ arcsin(x)/sqrt(1-x^2) dx=(arcsin(x))²- ∫ arcsin(x)/sqrt(1-x^2) dx =>2 ∫ arcsin(x)/sqrt(1-x^2) dx=(arcsin(x))²=>∫ arcsin(x)/sqrt(1-x^2) dx=(arcsin(x))²/2.