Single Variable Calculus, Chapter 5, 5.5, Section 5.5, Problem 34

Find the indefinite integral $\displaystyle \int \tan^2 \theta \sec ^2 \theta d \theta$. Illustrate by graphing both function and its anti-derivative (take $C = 0$).

If we let $u = \tan \theta$, then $\displaystyle du = \sec^2 \theta d \theta$. And


$
\begin{equation}
\begin{aligned}

\int \tan^2 \theta \sec ^2 \theta d \theta =& \int u^2 du
\\
\\
\int \tan^2 \theta \sec ^2 \theta d \theta =& \frac{u^{2 + 1}}{2 + 1} + C
\\
\\
\int \tan^2 \theta \sec ^2 \theta d \theta =& \frac{u^3}{3} + C
\\
\\
\int \tan^2 \theta \sec ^2 \theta d \theta =& \frac{(\tan \theta)^3}{3} + C
\\
\\
\int \tan^2 \theta \sec ^2 \theta d \theta =& \frac{\tan^3 \theta}{3} + C

\end{aligned}
\end{equation}
$


The graph of function $\displaystyle y' = \tan ^2 \theta \sec^2 \theta$








The graph of anti-derivative $\displaystyle y = \frac{\tan^3 \theta}{3}$

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