Answer
\[\frac{π}{16}\]
Work Step by Step
Let \[I=\int_{0}^{4}\frac{1}{16+t^2}dt\]
\[I=\int_{0}^{4}\frac{1}{t^2+(4)^2}dt\;\;\;...(1)\]
We will use the formula \[\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\;\;\;...(2)\]
By using (2) in (1)
\[\Rightarrow I=\left[\frac{1}{4}\tan^{-1}\left(\frac{x}{4}\right)\right]_{0}^{4}\]
\[\Rightarrow I=\frac{1}{4}\left[\tan^{-1}(1)-\tan^{-1}(0)\right]\]
\[\Rightarrow I=\frac{π}{16}\]
Hence , \[\int_{0}^{4}\frac{1}{16+t^2}dt=\frac{π}{16}\]
