Question

need help integrating this.
(arcsin(x))(1-x^2)^-1/2

Answer 1

Here is the original differential equation. Maybe I seperated the variables incorrectly.

y^2(1-x^2)^1/2*dy = arcsinx*dx

Answer 2

1/2(sin^-1(x))^2

Answer 3

see what you have here once you reconize it, is the D/dx of sin^-1(x) sitting right there
\[\int\limits_{}^{}\sin^-1(x)d(\sin^-1(x))=\int\limits_{}^{}\sin^-1(x)/\sqrt{1-x^2}dx\]

so because d(sin^-1(x))is sitting right there, you can take 1/2(sin^-1(x))^2 now because you know you need to take the derivative of sin^-1(x) again.

This is a technique not taught in many calc classes.

INT ln(x)/xdx=INT ln(x)d(ln(x))=1/2(ln(x))^2+C, but hopefully yhis was clear on my logic.