Definite integral of $y=\sqrt{(16-x^2)}$ The integral is 
$$\int^3_1\sqrt{16-x^2}dx$$
I've used the trig substitution method, replacing $x$ with $4\sin\theta$:
$$x=4\sin\theta, \quad \theta=\arcsin\left(\frac x4\right), \quad dx=4\cos\theta \ d\theta$$
(I've excluded the intervals for the definite integral for now)
\begin{align}
I&=\int\sqrt{16-16\sin^2\theta} \ 4\cos\theta  d\theta \\      
&=\int\sqrt{16\cos^2\theta}\ 4\cos\theta  d\theta\\
&=\int4\cos\theta 4\cos\theta d\theta \\
&=16\int\cos^2\theta d\theta \\
&=16\int\frac{\cos2\theta}{2}d\theta \\
&=8\int \cos2\theta d\theta\\
&=\left[4\sin2\theta\right]^{\arcsin(3/4)}_{\arcsin(1/4)}\\
&=4\sin\left(2\arcsin\frac 34\right)-4\sin\left(2\arcsin\frac 14\right)\\
\end{align}
At this last step, I'm not sure how to simplify the 2 arcsin part. It doesn't simplify because of the 2 in front of the arcsin, is there any other way? Or did I just do the wrong method to solve this integral in the first place?

$\sin(2\sin^{-1}(3/4))=2\sin(\sin^{-1}(3/4))\cos(\sin^{-1}(3/4))=2\cdot(3/4)\cdot(\sqrt{7}/4)$ because of the identity that $\cos(\sin^{-1}(3/4))=\sqrt{1-\sin^{2}(\sin^{-1}(3/4))}=\sqrt{1-9/16}=\sqrt{7}/4$.
