Is Trig Substitution Needed for This Integral?

[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

 

[Mark44]

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.

 

[Bashyboy]

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