# Trig Substitution

Bashyboy

## Homework Statement

$\int\frac{1}{\sqrt{16-x^2}}dx$

## Homework Equations

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

$4cos\theta=x$

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

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

## 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

Mentor

## Homework Statement

$\int\frac{1}{\sqrt{16-x^2}}dx$

## Homework Equations

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

$4cos\theta=x$

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

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

## 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

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##