Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 32

Determine the integral $\displaystyle \int \tan^6 (ay) dy$

Let $u = ay$, then $du = ady$, so $\displaystyle dy = \frac{du}{a}$. Thus,


$
\begin{equation}
\begin{aligned}

\int \tan^6 (ay) dy =& \int \tan^6 u \cdot \frac{du}{a}
\\
\\
\int \tan^6 (ay) dy =& \frac{1}{a} \int \tan^6 u du
\\
\\
\int \tan^6 (ay) dy =& \frac{1}{a} \int \tan^4 u \tan^2 u du
\qquad \text{Apply Trigonometric Identity } \sec^2 u = \tan^2 u + 1
\\
\\
\int \tan^6 (ay) dy =& \frac{1}{a} \int \tan^4 u (\sec^2 u - 1) du
\\
\\
\int \tan^6 (ay) dy =& \frac{1}{a} \int (\tan^4 u \sec^2 u - \tan^4 u) du
\\
\\
\int \tan^6 (ay) dy =& \frac{1}{a} \int \tan^4 u \sec^2 u du - \frac{1}{a} \int \tan^4u du

\end{aligned}
\end{equation}
$


We integrate the equation term by term

@ 1st term

$\frac{1}{a} \int \tan^4 u \sec^2 u du$

Let $v = \tan u$, then $dv = \sec^2 u du$. Thus


$
\begin{equation}
\begin{aligned}

\frac{1}{a} \int \tan^4 u \sec^2 u du =& \frac{1}{a} \int v^4 dv
\\
\\
\frac{1}{a} \int \tan^4 u \sec^2 u du =& \frac{1}{a} \left( \frac{v^{4 + 1}}{4 + 1} \right) + c
\\
\\
\frac{1}{a} \int \tan^4 u \sec^2 u du =& \frac{v^5}{5a} + c
\\
\\
\frac{1}{a} \int \tan^4 u \sec^2 u du =& \frac{(\tan u)^5}{5a} + c
\\
\\
\frac{1}{a} \int \tan^4 u \sec^2 u du =& \frac{\tan^5 u}{5a} + c


\end{aligned}
\end{equation}
$


@ 2nd term



$
\begin{equation}
\begin{aligned}

\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int \tan^2 u \tan^2 u du
\qquad \text{Apply Trigonometric Identity } \sec^2 u = \tan^2 u + 1
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int \tan^2 u (\sec^2 u - 1) du
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int (\tan^2 u \sec^2 u - \tan^2 u) du
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int \tan^2 u \sec^2 u du - \frac{1}{a} \int \tan^2 u du
\qquad \text{Apply Trigonometric Identity } \sec^2 u = \tan^2 u + 1 \text{ for the 2nd term}
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int \tan^2 u \sec^2 u du - \frac{1}{a} \int (\sec^2 u - 1) du

\end{aligned}
\end{equation}
$


For the 1st term, let $v = \tan u$, then $dv = \sec^2 u du$. Thus,


$
\begin{equation}
\begin{aligned}

\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \int v^2 dv - \frac{1}{a} (\tan u - u) + c
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{1}{a} \left( \frac{v^{2 + 1}}{2 + 1}\right) - \frac{1}{a} ( \tan u - u ) + c
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{v^3}{3a} - \frac{\tan u}{a} + \frac{u}{a} + c
\qquad \text{Substitute value of } v
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{(\tan u)^3}{3a} - \frac{\tan u}{a} + \frac{u}{a} + c
\\
\\
\frac{1}{a} \int \tan^4 u du =& \frac{\tan^3 u}{3a} - \frac{\tan u}{a} + \frac{u}{a} + c

\end{aligned}
\end{equation}
$


Combine the results of the integration term by term


$
\begin{equation}
\begin{aligned}

\int \tan^6 (ay) dy =& \frac{\tan^5 u}{5a} - \left( \frac{\tan^3u}{3a} -\frac{\tan u}{a} - \frac{\tan u}{a} + \frac{u}{a} \right) + c
\\
\\
\int \tan^6 (ay) dy =& \frac{\tan^5 u}{5a} - \frac{\tan^3 u}{3a} + \frac{\tan u}{a} - \frac{u}{a} + c
\qquad \text{Substitute value of } u
\\
\\
\int \tan^6 (ay) dy =& \frac{\tan^5 (ay)}{5a} - \frac{\tan^3 (ay)}{3a} + \frac{\tan (ay)}{a} - \frac{\cancel{a}y}{\cancel{a}} + c
\\
\\
\int \tan^6 (ay) dy =& \frac{\tan^5 (ay)}{5a} - \frac{\tan^3 (ay)}{3a} + \frac{\tan (ay)}{a} - y + c

\end{aligned}
\end{equation}
$