Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 92

Evaluate $\displaystyle \int^4_0 \frac{1}{16 + t^2} dt$

If we let $t = 4u$, then $dt = 4du$

Make sure that the upper and lower limits are also in terms of $u$

So,


$
\begin{equation}
\begin{aligned}

\int^4_0 \frac{1}{16 + t^2} dt =& \int^{\frac{4}{4}}_{\frac{0}{4}} \frac{4 du }{16 + (4u)^2}
\\
\\
=& \int^1_0 \frac{4du}{16 + 16u^2}
\\
\\
=& \int^1_0 \frac{4du}{16 (1 + u^2)}
\\
\\
=& \frac{1}{4} \int^1_0 \frac{du}{1 + u^2}

\end{aligned}
\end{equation}
$


Recall that

$\displaystyle \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2}$

Thus,


$
\begin{equation}
\begin{aligned}

=& \frac{1}{4} \left[ \tan^{-1} x \right]^1_0
\\
\\
=& \frac{1}{4} \left[ \tan^{-1} (1) - \tan^{-1} (0) \right]^1_0
\\
\\
=& \frac{1}{4} \tan^{-1} (1)
\\
\\
=& \frac{\pi}{16}

\end{aligned}
\end{equation}
$