Sine substitution? My book says the following:
$$\int \frac{dx}{(16-x^2)^{3/2}}$$ 
$$x = 4\sin\theta$$
$$(16 - x^2)^{3/2} = (4^2\cos^2\theta)^{3/2}$$
$$=(4\cos\theta)^3$$
I don't understand the last step:
Doesn't:
$$(4^2\cos^2\theta)^{3/2} = (|4\cos\theta|)^3$$
Since:
$\sqrt{x^2} = |x|$

We are really letting $\theta=\arcsin(x/4)$. So $\theta$ ranges over the interval $(-\pi/2,\pi/2)$, and the cosine in this interval is non-negative. 
