# Trig Substitution

1. Sep 15, 2012

### Bashyboy

1. The problem statement, all variables and given/known data
$\int\frac{1}{\sqrt{16-x^2}}dx$

2. Relevant equations
$csc\theta=\frac{4}{\sqrt{16-x^2}}$

$4cos\theta=x$

$-4sin\theta d\theta=dx$

$\theta=arccos(\frac{x}{4})$

3. The attempt at a solution
Using these facts, I concluded that the integral, after all of the substitution, was
$-arccos(\frac{x}{4})$ but the actual answer is the arcsin

2. Sep 15, 2012

### Staff: Mentor

You don't show your complete answer, but what I think you have is probably correct. There is an identity that involves inverse trig functions:
arcsin(x) + arccos(x) = $\pi/2$

So arcsin(x) = -arccos(x) + $\pi/2$