Calculus (3rd Edition)

$$\frac{8}{\sqrt{5}}(\sin^{-1}(\frac{\sqrt{5} x}{4})+\frac{\sqrt{5} x}{4}\frac{\sqrt{16-5x^2}}{4}) +c$$
Since $x=\frac{4}{\sqrt{5}}\sin \theta$, then $dx=\frac{4}{\sqrt{5}}\cos \theta d \theta$ and hence $$\int \sqrt{16-5x^2} dx=\frac{4}{\sqrt{5}}\int \sqrt{16-16 \sin^2 \theta} \cos \theta d \theta\\ =\frac{16}{\sqrt{5}}\int \cos^2 \theta d \theta\\ =\frac{16}{\sqrt{5}}\int \cos^2 \theta d \theta\\ =\frac{8}{\sqrt{5}}\int 1+ \cos2 \theta d \theta\\ =\frac{8}{\sqrt{5}}(\theta+\frac{1}{2} \sin2 \theta )+c\\ =\frac{8}{\sqrt{5}}(\theta+ \sin \theta \cos \theta)+c\\ =\frac{8}{\sqrt{5}}(\sin^{-1}(\frac{\sqrt{5} x}{4})+\frac{\sqrt{5} x}{4}\frac{\sqrt{16-5x^2}}{4}) +c$$ Note that we used the identity $\sin 2\theta=2\cos \theta \sin \theta$.