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
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\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
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\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
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\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
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\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
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\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}
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\int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{1}{7} + \frac{1}{5}
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\int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{5 + 7}{35}
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\int^{\frac{\pi}{4}}_0 (\tan^6 \theta + \tan^4 \theta) \sec^2 \theta d \theta =& \frac{12}{35}

\end{aligned}
\end{equation}
$