<s>[INST] Evaluating $\int \sqrt{16-x^2} \,dx$ Are the steps below correct? Can you tell me another way to solve it, just hint?
$   \int \sqrt{16-x^2}   \,dx $
$ x = 4 \sin u \rightarrow  u=\arcsin(\frac{x}{4}) \rightarrow  dx = 4\cos u \,dx$
$   \int  \sqrt{16-x^2}   \,dx   = 16 \int \cos^2 u \,du = 16 \int \frac{\cos 2u +1}{2} \,du  $
$ v = 2u  \Rightarrow  dv = 2 du \Rightarrow du = \frac{1}{2}dv  $
$  16 \int \frac{\cos 2u +1}{2} \,du = 16 \int \frac{\cos v}{2} +\frac{1}{2} \,dv = 16(\frac{\sin v}{2} +\frac{v}{2}) = 8\sin v +8v = 8\sin 2u +16u  $
$ \Downarrow $
$8\sin(2\arcsin(\frac{x}{4})) + 4\arcsin\frac{x}{4} $
 [/INST] Let $I = \int\sqrt{16-x^2} \ dx$. Then applying integration by parts with $u = \sqrt{16-x^2}, du = -\frac{x}{\sqrt{16-x^2}}$ and $dv = 1, v = x$:
$$I = x \sqrt{16-x^2} + \int \frac{x}{\sqrt{16-x^2}} \ dx$$
and now let $w = 16-x^2$.
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