Single Variable Calculus, Chapter 8, 8.4, Section 8.4, Problem 52

Find the indefinite integral $\displaystyle \int \sec^4 \left( \frac{x}{2} \right) dx$. Illustrate by graphing both the integrand and its antiderivative (taking $c = 0$).

Let $\displaystyle z = \frac{x}{2}$, then $\displaystyle dz = \frac{1}{2} dx$, so $dx = 2dz$. Thus,

$
\begin{equation}
\begin{aligned}
\int sec^4 \left( \frac{x}{2} \right) dx &= \int \sec^4 z \cdot 2 dz\\
\\
\int sec^4 \left( \frac{x}{2} \right) dx &= 2 \int \sec^4 z dz\\
\\
&= 2 \int \sec^2 z \cdot \sec^2 z dz
\end{aligned}
\end{equation}
$


By using Integration by parts...

If we let $u = \sec^2 z$, then $du = 2 \sec z(\sec z \tan z) dz$ and if $dv = \sec^2 z dz$, then $\displaystyle v = \int \sec^2 z dz = \tan z$
So,

$
\begin{equation}
\begin{aligned}
\int \sec^2 z - \sec^2 z dz &= uv - \int v du = \sec^2 z \tan z - \int 2 (\tan z) \sec z (\sec z \tan z) dz\\
\\
&= \sec^2 z \tan z - \int 2 \sec^2 z \tan^2 z dz && \text{recall that } \tan^2 z = \sec^2 z - 1\\
\\
&= \sec^2 z \tan z - 2 \int \sec^2 z (\sec^2 z - 1) dz\\
\\
&= \sec^2 z \tan z - 2 \int \sec^4 z dz + 2 \int \sec^2 z dz
\end{aligned}
\end{equation}
$


By combining like terms,

$
\begin{equation}
\begin{aligned}
\int \sec^4 z dz &= \sec^2 z \tan z - 2 \int \sec^4 z dz + 2 \tan z\\
\\
\int \sec^4 z dz + 2 \int \sec^4 zdz &= \sec^2 z \tan z + 2 \tan z + c\\
\\
3 \int \sec^4 z dz &= \sec^2 z \tan z + 2 \tan z + c\\
\\
\int \sec^4 z dz &= \frac{\sec^2 z \tan z + 2 \tan z}{3} + c\\
\\
2\int \sec^4 z dz &= 2 \left[ \frac{\sec^z \tan z + z \tan z}{3} \right] +c\\
\\
2\int \sec^4 z dz &= \frac{2}{3} \left[ \sec^2 z \tan z + 2 \tan z \right] + c

\end{aligned}
\end{equation}
$


But $\displaystyle z = \frac{x}{2}$, so
$\displaystyle \int \sec^4 \left( \frac{x}{2} \right) dx = \frac{2}{3} \left[ \sec^2 \left( \frac{x}{2} \right) \tan \left( \frac{x}{2} \right) + 2 \tan \left( \frac{x}{2} \right) \right] $