# Trig Substitution

## 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

## Answers and Replies

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## 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##

Your answer (-arccos(x/4)) and the book's answer (arcsin(x/4)) differ by a constant.

Also, you can always check your answer to an integration problem by differentiating it. If the result is your original integrand, then your answer is correct.

Oh, okay. It certainly makes perfect sense that there would be a inverse trig identity. Thank you.