Dtyjlui

One can easily show that $intlimits_0^{pi}frac{cos(2k+1)x}{cos x} dx = 2 intlimits_0^{frac{pi}{2}}frac{cos(2k+1)x}{cos x} dx = (-1)^k pi$.

But is there a closed form for $intlimits_0^{frac{pi}{4}}frac{cos(2k+1)x}{cos x} dx$ ?

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
$$ frac{cos((2k+1) x)}{cos(x)} = mathrm{e}^{2 k mathrm{i} x} frac{1+mathrm{e}^{-(2k+1) 2 mathrm{i} x}}{1+mathrm{e}^{-2mathrm{i}x}} = sum limits_{l=0}^{2k} (-1)^l mathrm{e}^{2 (k-l) mathrm{i} x} = (-1)^k left[1+2sum limits_{m=1}^k (-1)^m cos(2mx)right]$$
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} sinleft(frac{mpi}{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$ .

It is worthwhile to recall that for $ninBbb N$,
$$cos nx=T_n(cos x)$$
Where $T_n(x)$ is the Chebyshev polynomial of the first kind, defined as
$$T_n(x)=frac{n}2sum_{r=0}^{lfloor n/2rfloor}frac{(-1)^r}{n-r}{n-rchoose r}(2x)^{n-2r}$$
Hence
$$I_k=int_0^{pi/4}frac{cos[(2k+1)x]}{cos x}dx$$
$$I_k=int_0^{pi/4}frac{2k+1}{2cos x}sum_{r=0}^{lfloor frac{2k+1}2rfloor}frac{(-1)^r}{2k+1-r}{2k+1-rchoose r}(2cos x)^{2k-2r+1}dx$$
$$I_k=frac{2k+1}{2}sum_{r=0}^{lfloor frac{2k+1}2rfloor}frac{(-1)^r}{2k+1-r}{2k+1-rchoose r}2^{2k-2r+1}int_0^{pi/4}cos(x)^{2k-2r}dx$$
$$I_k=(2k+1)sum_{r=0}^{k}frac{(-1)^r4^{k-r}}{2k-r+1}{2k-r+1choose r}int_0^{pi/4}cos(x)^{2k-2r}dx$$
Then we focus on the integral
$$J(r,k)=int_0^{pi/4}cos(x)^{2(k-r)}dx$$
Unfortunately, the closest to a closed form for this guy is something in terms of the incomplete Beta function. Here's how:

Consider the integral
$$H(k;a,b)=int_0^ksin(x)^{a}cos(x)^{b}dt$$
For some $0leq kleq 1$. Making the substitution $t=sin(x)^2$, we see that
$$H(k;a,b)=frac12int_0^{sin(k)^2}t^{frac{a-1}2}(1-t)^{frac{b-1}2}dt$$
$$H(k;a,b)=frac12int_0^{sin(k)^2}t^{frac{a+1}2-1}(1-t)^{frac{b+1}2-1}dt$$
Then Recall the definition of the incomplete Beta function:
$$mathrm{B}(x;a,b)=int_0^x t^{a-1}(1-t)^{b-1}dt$$
Thus
$$H(k;a,b)=frac12mathrm{B}bigg(sin(k)^2;frac{a+1}2,frac{b+1}2bigg)$$
And $$J(r,k)=H(pi/4;0,2k-2r)=frac12mathrm{B}bigg(frac12;frac{1}2,k-r+frac{1}2bigg)$$
Thus
$$I_k=frac{2k+1}2sum_{r=0}^{k}frac{(-1)^r4^{k-r}}{2k-r+1}{2k-r+1choose r}mathrm{B}bigg(frac12;frac{1}2,k-r+frac{1}2bigg)$$
Which could be considered a closed form.