Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 26

Determine the integral $\displaystyle \int^{\frac{\pi}{4}}_0 \sec^4 \theta \tan^4 \theta d \theta$



$\begin{equation} \begin{aligned} \int^{\frac{\pi}{4}}_0 \sec^4 \theta \tan^4 \theta d \theta =& \int^{\frac{\pi}{4}}_0 \sec^2 \theta \sec^2 \theta \tan^4 \theta d \theta \qquad \text{Apply Trigonometric Identity } \sec^2 \theta = \tan^2 \theta + 1 \text{ for } \sec^2 \theta \\ \\ \int^{\frac{\pi}{4}}_0 \sec^4 \theta \tan^4 \theta d \theta =& \int^{\frac{\pi}{4}}_0 (\tan^2 \theta + 1) \tan^4 \theta \sec^2 \theta d \theta \\ \\ \int^{\frac{\pi}{4}}_0 \sec^4 \theta \tan^4 \theta d \theta =& \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta \end{aligned} \end{equation}$


Let $u = \tan \theta$, then $du = \sec^2 \theta d \theta$. When $\displaystyle \theta = 0, u = 0, \theta = \frac{\pi}{4}, u =1$. Therefore,


$\begin{equation} \begin{aligned} \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \int^1_0 (u^6 + u^4) du \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \left[ \frac{u^{6 + 1}}{6 + 1} + \frac{u^{4 + 1}}{4 + 1} \right]^1_0 \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \left[ \frac{u^7}{7} + \frac{u^5}{5} \right]^1_0 \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{(1)^7}{7} + \frac{(1)^5}{5} - \frac{(0)^7}{7} - \frac{(0)^5}{5} \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{1}{7} + \frac{1}{5} \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{5 + 7}{35} \\ \\ \int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{12}{35} \end{aligned} \end{equation}$