Is there a closed form for the trigonometric integral $\int\limits_0^{\pi/4}\frac{\cos(2k+1)x}{\cos x} dx$? One can easily show that $\int\limits_0^{\pi}\frac{\cos(2k+1)x}{\cos x} dx = 2 \int\limits_0^{\frac{\pi}{2}}\frac{\cos(2k+1)x}{\cos x} dx  = (-1)^k \pi$.
But is there a closed form for $\int\limits_0^{\frac{\pi}{4}}\frac{\cos(2k+1)x}{\cos x} dx$ ?
So far, I tried by induction and by IBP but couldn't come to a simple closed form.
Thanks for your hints !

A sketch:
Write
$$ \frac{\cos((2k+1) x)}{\cos(x)} =  \mathrm{e}^{2 k \mathrm{i} x} \frac{1+\mathrm{e}^{-(2k+1) 2 \mathrm{i} x}}{1+\mathrm{e}^{-2\mathrm{i}x}} = \sum \limits_{l=0}^{2k} (-1)^l \mathrm{e}^{2 (k-l) \mathrm{i} x} = (-1)^k \left[1+2\sum \limits_{m=1}^k (-1)^m \cos(2mx)\right]$$
to obtain
\begin{align} I_k &\equiv \int \limits_0^{\pi/4} \frac{\cos((2k+1) x)}{\cos(x)} \, \mathrm{d} x = (-1)^k \left[\frac{\pi}{4} + \sum \limits_{m=1}^k \frac{(-1)^m}{m} \sin\left(\frac{m\pi}{2}\right)\right] \\
&= (-1)^k \left[\frac{\pi}{4} - \sum \limits_{n=0}^{\lfloor \frac{k-1}{2}\rfloor} \frac{(-1)^n}{2n+1}\right]
\end{align}
for $k \in \mathbb{N}_0$ in agreement with Crostul's comment. In particular, $ \lim_{k \to \infty} I_k = 0$ .
