Inverse Trigonometric Formulas

[JasonRox]

Someone has to help me here.

1. \int \frac{dx}{\sqrt{1 - x^2}} = sin^-1 x + C

So the derivative of sin^-1x is \frac{dx}{\sqrt{1 - x^2}}?

The -1 are exponents. It isn't working.

[Caldus]

The derivative of sin inverse is:

1/1 - x2)

[Njorl]

Do the integral with a trig substitution, x=sinu, dx=cosudu

Use the identity sin2u+cos2u=1, or 1-sin2u=cos2u

This way, 1-x2 becomes 1-sin2u becomes cos2u

Your integral reduces to INTdu=u+c=sin-1+c

Njorl