Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 50

Express the vlaue of $\displaystyle \int^{\pi/4}_0 \tan^8 x \sec x dx$ in terms of $I$. Suppose that $\displaystyle I = \int^{\pi/4}_0 \tan^6 x \sec x dx$.
By using integration by parts,
If we let $u = \tan^7 x$ and $dv = \tan x \sec x dx$, then
$du = 7 \tan^6 x \sec^2 x dx$ and $\displaystyle v = \int \tan x \sec x dx = \sec x$

So,

$
\begin{equation}
\begin{aligned}
\int^{\pi/4}_0 \tan^8 x \sec x dx &= \int^{\pi/4}_0 \tan^7 \cdot \tan x \cdot \sec x dx = uv - \int vdu\\
\\
&= \tan^7 x \sec x - \int \sec \left( 7 \tan ^6 x \sec^2 x dx \right)\\
\\
&= \tan^7 x \sec x - \int 7 \tan^6 x \sec^2 x \sec x dx
\end{aligned}
\end{equation}
$


Recall that $\sec^2 x = 1 + \tan^2 x$

$
\begin{equation}
\begin{aligned}
\phantom{\int^{\pi/4}_0 \tan^8 x \sec x dx}&= \tan^6 x \sec x - \int 7 \tan^6 x \left( 1 + \tan^2 x \right) \sec x dx\\
\\
&= \tan^6 \sec x - 7 \int \tan^6 x \sec x dx - 7 \int \tan^8 x \sec x dx
\end{aligned}
\end{equation}
$


By combining like terms

$
\begin{equation}
\begin{aligned}
\int \tan^8 x \sec x dx + 7 \int \tan^8 x \sec x dx &= \tan^7 x \sec x - 7 \int \tan^6 x \sec x dx\\
\\
8 \int \tan^8 x \sec x dx &= \tan^7 x \sec x -7 \int \tan^6 x \sec x dx\\
\\
\int \tan^8 x \sec x dx &= \frac{\tan^7 x \sec x -7 \int \tan^6 x \sec x dx }{8}
\end{aligned}
\end{equation}
$


Evaluating from 0 to $\displaystyle \frac{\pi}{4}$,
$\displaystyle = \left[ \frac{\tan^7 x \sec x}{8} \right]^{\pi/4}_0 - \frac{7}{8} \int^{\pi/4}_0 \tan^6 x \sec x dx$

but, $\displaystyle I = \int^{\pi/4}_0 \tan^6 x \sec x dx$, so

$
\begin{equation}
\begin{aligned}
\int^{\pi/4}_0 \tan^8 x \sec x dx &= \left[ \frac{\tan^7 \left( \frac{\pi}{4} \right) \sec \left( \frac{\pi}{4} \right) }{8} \right] - \left[ \frac{\tan^7(0) \sec(0)}{8} \right] - \frac{7}{8} I\\
\\
\int^{\pi/4}_0 \tan^8 x \sec x dx &= \frac{\sqrt{2}}{8} - \frac{7}{8} I = \frac{1}{8} (\sqrt{2} - 7 I)
\end{aligned}
\end{equation}
$