I don't see how / why one can rewrite the integral as following:
$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \,dt = \frac{2}{\frac{\pi}{2}+j\pi} $$
I think this should be rather easy, but I don't see what I'm missing.
Because
$$\int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} |\cos(t)| \,dt = 2 $$ for any $j \in \mathbb Z$.
HINT:
Note that $\int_{(j-1)\pi+\pi/2}^{j\pi+\pi/2}|\cos(x) |\,dx=2$. And the denominator of the integrand is a constant with respect to the variable of integration.
Get the denominator outside the integral, as it's a constant.
You are integrating $|\cos(x)|$, which is $\pi$-periodic, on an interval on length $\pi$. The result is the same as
$$\int_{-\pi/2}^{\pi/2}\cos(x)dx=[\sin x]_{-\pi/2}^{\pi/2}=2$$
We can write \begin{align*} \int_{\frac{\pi}{2}+(j-1)\pi}^{\frac{\pi}{2}+j\pi} \frac{|\cos(t)|}{\frac{\pi}{2}+j\pi} \ dt & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{\frac{-\pi}{2}+j\pi}^{\frac{\pi}{2}+j\pi}|\cos(t)| \ dt\\ &\overset{\color{blue}{u=t-j\pi} }{=} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du\\ \end{align*} and if $j$ is an integer, we know that $$|\cos(u + j\pi)| = |(-1)^j \cos(u)| = |\cos(u)| $$ so we get\begin{align*} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u+j\pi)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du \end{align*} and on the interval $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ we know that cosine is positive, so on this interval (which is the one we're integrating on) we get $|\cos(u)| = \cos(u)$. And lastly we see \begin{align*} \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}|\cos(u)| \ du & = \frac{1}{\frac{\pi}{2}+j\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(u) \ du\\ & = \frac{1}{\frac{\pi}{2}+j\pi} \sin(u)\Biggr|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\\ & = \frac{1}{\frac{\pi}{2}+j\pi} \left(1 - (-1)\right)\\ & = \frac{2}{\frac{\pi}{2}+j\pi} \end{align*}
