Answer
$$ \int \sec ^{4}(2 \pi x / 5) d x =\frac{5}{2\pi }\left(\frac{1}{3} \sec ^{2} (2 \pi x / 5) \tan (2 \pi x / 5)+\frac{2}{3} \tan (2 \pi x / 5) \right)+C$$
Work Step by Step
$$
\int \sec ^{4}(2 \pi x / 5) d x
$$
Let $ u=2 \pi x / 5\ \ \to\ \ \ (5/2\pi)du= dx $
$$
\int \sec ^{4}(2 \pi x / 5) d x=\frac{5}{2\pi } \int \sec ^{4}(u) d u
$$
Here $n= 4$, Use the formula $$
\int \sec ^{n} x d x=\frac{1}{n-1} \sec ^{n-2} x \tan x+\frac{n-2}{n-1} \int \sec ^{n-2} x d x
$$
Then
\begin{align*}
\int \sec ^{4}(2 \pi x / 5) d x&=\frac{5}{2\pi } \int \sec ^{4}(u) d u\\
&=\frac{5}{2\pi }\left(\frac{1}{3} \sec ^{2} u \tan u+\frac{2}{3} \int \sec ^{2} u d u\right)\\
&=\frac{5}{2\pi }\left(\frac{1}{3} \sec ^{2} u \tan u+\frac{2}{3} \tan u \right)+C\\
&=\frac{5}{2\pi }\left(\frac{1}{3} \sec ^{2} (2 \pi x / 5) \tan (2 \pi x / 5)+\frac{2}{3} \tan (2 \pi x / 5) \right)+C
\end{align*}
