Substitute [tex]x = 4 \sin(y)[/tex], so that [tex]dx = 4\cos(y)\,dy[/tex]. Part of the integrand reduces to
[tex]16 - x^2 = 16 - (4\sin(y))^2 = 16 - 16 \sin^2(y) = 16 (1 - \sin^2(y)) = 16 \cos^2(y)[/tex]
Note that we want this substitution to be reversible, so we tacitly assume [tex]-\frac\pi2\le y\le \frac\pi2[/tex]. Then [tex]\cos(y)\ge0[/tex], and
[tex](16-x^2)^{3/2} = 16^{3/2} \left(\cos^2(y)\right)^{3/2} = 64 |\cos(y)|^3 = 64 \cos^3(y)[/tex]
(since [tex]\sqrt{x^2} = |x|[/tex] for all real [tex]x[/tex])
So, the integral we want transforms to
[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx = 64 \int \cos^3(y) \times 4\cos(y) \, dy = 256 \int \cos^4(y) \, dy[/tex]
Expand the integrand using the identity
[tex]\cos^2(x) = \dfrac{1+\cos(2x)}2[/tex]
to write
[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx = 256 \int \left(\frac{1 + \cos(2y)}2\right)^2 \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \cos^2(2y)) \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \frac{1 + \cos(4y)}2\right) \, dy \\\\ = 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy[/tex]
Now integrate to get
[tex]\displaystyle 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy = 32 \left(3y + 2 \sin(2y) + \frac14 \sin(4y)\right) + C \\\\ = 96 y + 64 \sin(2y) + 8 \sin(4y) + C[/tex]
Recall the double angle identity,
[tex]\sin(2y) = 2 \sin(y) \cos(y)[/tex]
[tex]\implies \sin(4y) = 2 \sin(2y) \cos(2y) = 4 \sin(y) \cos(y) (\cos^2(y) - \sin^2(y))[/tex]
By the Pythagorean identity,
[tex]\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \dfrac{x^2}{16}} = \dfrac{\sqrt{16-x^2}}4[/tex]
Finally, put the result back in terms of [tex]x[/tex].
[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx \\\\ = 96 \sin^{-1}\left(\frac x4\right) + 128 \frac x4 \frac{\sqrt{16-x^2}}4 + 32 \frac x4 \frac{\sqrt{16-x^2}}4 \left(\frac{16-x^2}{16} - \frac{x^2}{16}\right) + C \\\\ = 96 \sin^{-1}\left(\frac x4\right) + 8 x \sqrt{16 - x^2} + \frac14 x \sqrt{16 - x^2} (8 - x^2) + C \\\\ = \boxed{96 \sin^{-1}\left(\frac x4\right) + \frac14 x \sqrt{16 - x^2} \left(40 - x^2\right) + C}[/tex]
