Is it possible to find a explicit formula here ?
This is the reduction formula of$$\int_{0}^{\frac{\pi}{4}} (\cos x)^n dx$$.
If the limits are from $0$ to $\frac{\pi}{2}$ then the nice walli's formula can be obtained,however i have no idea regarding solving this kind of recurrences. Any help will be appreciated.
On
Starting with $$ y_n=\frac1{n\,2^{n/2}}+\frac{n-1}{n} y_{n-2}\tag1 $$ Multiply $(1)$ by $\frac{\Gamma\left(\frac{n+2}2\right)}{\Gamma\left(\frac{n+1}2\right)}=\frac{n}2\frac{\Gamma\!\left(\frac{n}2\right)}{\Gamma\!\left(\frac{n+1}2\right)}=\frac{n}{n-1}\frac{\Gamma\left(\frac{n}2\right)}{\Gamma\left(\frac{n-1}2\right)}$: $$ \frac{\Gamma\!\left(\frac{n+2}2\right)}{\Gamma\!\left(\frac{n+1}2\right)}y_n =\frac1{2^{\frac{n}2+1}}\frac{\Gamma\!\left(\frac{n}2\right)}{\Gamma\!\left(\frac{n+1}2\right)}+\frac{\Gamma\!\left(\frac{n}2\right)}{\Gamma\!\left(\frac{n-1}2\right)} y_{n-2}\tag2 $$ Thus, we get $$ \overbrace{\frac{\Gamma\!\left(n+1\right)}{\Gamma\!\left(n+\frac12\right)}y_{2n}}^{\large\alpha_n} =\frac1{2^{n+1}}\frac{\Gamma\!\left(n\right)}{\Gamma\!\left(n+\frac12\right)}+\overbrace{\frac{\Gamma\!\left(n\right)}{\Gamma\!\left(n-\frac12\right)}y_{2n-2}}^{\large\alpha_{n-1}}\tag3 $$ and $$ \overbrace{\frac{\Gamma\!\left(n+\frac32\right)}{\Gamma\!\left(n+1\right)}y_{2n+1}}^{\large\beta_n} =\frac1{2^{n+\frac32}}\frac{\Gamma\!\left(n+\frac12\right)}{\Gamma\!\left(n+1\right)}+\overbrace{\frac{\Gamma\!\left(n+\frac12\right)}{\Gamma\!\left(n\right)}y_{2n-1}}^{\large\beta_{n-1}}\tag4 $$ Therefore, noting that $\alpha_n=\alpha_0+\sum\limits_{k=1}^n(\alpha_k-\alpha_{k-1})$, $$ y_{2n}=\frac{\Gamma\!\left(n+\frac12\right)}{\Gamma\!\left(n+1\right)}\left[\frac{y_0}{\sqrt\pi}+\sum_{k=1}^n\frac1{2^{k+1}}\frac{\Gamma\!\left(k\right)}{\Gamma\!\left(k+\frac12\right)}\right]\tag5 $$ and noting that $\beta_n=\beta_0+\sum\limits_{k=1}^n(\beta_k-\beta_{k-1})$, $$ y_{2n+1}=\frac{\Gamma\!\left(n+1\right)}{\Gamma\!\left(n+\frac32\right)}\left[\frac{y_1\sqrt\pi}2+\sum_{k=1}^n\frac1{2^{k+\frac32}}\frac{\Gamma\!\left(k+\frac12\right)}{\Gamma\!\left(k+1\right)}\right]\tag6 $$ It might be useful to recall that $\Gamma(n+1)=n!$ and $\Gamma\!\left(n+\tfrac12\right)=n!\frac{\sqrt\pi}{4^n}\binom{2n}{n}$.
Unfortunately, I am not aware of closed forms for the sums in $(5)$ and $(6)$.
On
But why not use the following relationship
$$\cos^nx=\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}\cos(n-2k)x$$
Then
$$y_{n}=\int_{0}^{\frac{\pi}{4}}\cos^nx \ dx=\frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}\frac{\sin\frac{(n-2k)\pi}{4}}{n-2k}$$
The last formula works for $n=2m+1$
For $n=2m$ we get
$$y_{n}=\int_{0}^{\frac{\pi}{4}}\cos^nx \ dx=\frac{\pi}{4}\frac{\binom{n}{\frac{n}{2}}}{2^n}+\frac{1}{2^{n-1}}\sum_{k=0}^{\frac{n}{2}-1}\binom{n}{k}\frac{\sin\frac{(n-2k)\pi}{4}}{n-2k}$$
We can find an explicit formula for $y_n$ when $n$ is even. Here's how. Define $$D_n=y_{2n}$$ Hence we have $$D_n=\frac1{2n\cdot (\sqrt2)^{2n}}+\frac{2n-1}{2n}D_{n-1}$$ $$D_n=\frac1{2^{n+1}n}+\frac{2n-1}{2n}D_{n-1}$$ And we have the base case $D_0=\int_0^{\pi/4}dx=\pi/4$.
Then we note that, given the general recurrence $$f_n=\alpha_n+\beta_nf_{n-1}$$ Where $\alpha, \beta$ are explicit functions of $n$ and the base case $f_0$ is known. We have $$f_n=f_0\prod_{i=1}^{n}\beta_i+\sum_{k=0}^{n-1}\alpha_{n-k}\prod_{j=1}^{k}\beta_{n-j+1}$$ Which is an explicit function of $n$. Choosing $\alpha_n=\frac1{2^{n+1}n}$ and $\beta_n=\frac{2n-1}{2n}$ We have $$D_n=\frac\pi4\prod_{i=1}^{n}\frac{2i-1}{2i}+\sum_{k=0}^{n-1}\frac{2^{k-n-1}}{n-k}\prod_{j=1}^{k}\frac{2n-2j+1}{2n-2j+2}$$ $$D_n=\frac\pi{4^{n+1}}{2n\choose n}+\frac1{2^{n+1}}\sum_{k=0}^{n-1}\frac{1}{n-k}\prod_{j=1}^{k}\frac{2n-2j+1}{n-j+1}$$ I do not know how to simplify the final product anymore though. Note that this formula for $D_n$ works for integer $n\geq0$ because $\sum_{k=0}^{-1}:=0$ and $\prod_{j=1}^{0}:=1$
I hope this helped :)