Answer
$$\frac{1}{16} \frac{t}{\sqrt{16-t^{2}}}+C$$
Work Step by Step
\begin{aligned}
\int \frac{d t}{\left(16-t^{2}\right)^{\frac{3}{2}}} &=\int \frac{4 \cos \theta d \theta}{\left(16-(4 \sin \theta)^{2}\right)^{\frac{3}{2}}} \\
&=\int \frac{4 \cos \theta d \theta}{(\sqrt{16-16 \sin ^{2} \theta})^{3}} \\
&=\int \frac{4 \cos \theta d \theta}{(4 \sqrt{1-\sin ^{2} \theta})^{3}} \\
&=\frac{1}{16} \int \frac{\cos \theta d \theta}{(\sqrt{\cos ^{2} \theta})^{3}} \\
&=\frac{1}{16} \int \frac{\cos \theta d \theta}{(\cos \theta)^{3}}\\
&=\frac{1}{16}\int \sec^2 \theta d\theta\\
&=\frac{1}{16}\tan \theta +C\\
&=\frac{1}{16} \frac{t}{\sqrt{16-t^{2}}}+C
\end{aligned}
