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/[square]1 - x²[/square])

 

[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