# 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{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}

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

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

Posted on