Math Challenges

Solution:
\( \displaylines{\text{Using }x=4\sin\left(\theta\right),\differentialD x=4\cos\left(\theta\right)d\theta\implies\int_{}^{}\frac{\sqrt{\left(16-x^2\right)}}{x^2}dx\\ =\int_{}^{}\frac{\cos^2\theta}{\sin^2\theta}d\theta=\int_{}^{}\frac{1-\sin^2\left(\theta\right)}{\sin^2\left(\theta\right)}d\theta=\int_{}^{}\left(\csc^2\left(\theta\right)-1\right)d\theta\\ =-\cot\left(\theta\right)-\theta+C\\ =-\frac{\sqrt{\left(16-x^2\right)}}{x}-\arcsin\left(\frac{x}{4}\right)+C} \)