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

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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{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}

Recall that

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

Thus,

\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}