Cartwright's proof that $\pi$ is irrational induction part I am currently proving that $\pi$ is irrational, and I am basing from Cartwright's proof. I let
\begin{equation*}
    A_k(x) = \int_{-1}^{1} (1-y^2)^k \cos(xy)dy
\end{equation*}
where $k \in \{0\} \cup \mathbb{Z}^+$. I then used integration by parts and got
\begin{equation*}
    x^2A_k(x) = 2k(2k-1)A_{k-1}(x) - 4k(k-1)A_{k-2}(x), k \geq 2.
\end{equation*}
Next, I let
$B_k(x) = x^{2k+1}A_k(x)$ for all nonnegative integer $k$.
I have already shown
\begin{align*}
B_k(x) &= 2k(2k-1)B_{k-1}(x) - 4k(k-1)x^2B_{k-2}(x)\\
B_0(x) &= 2\sin x\\
B_1(x) &= 4\sin x -4x\cos x\\
\end{align*}
I am struggling to show
\begin{equation}
    B_k(x) = k!(M_k(x)\sin x + N_k(x)\cos x)
\end{equation}
where $M_k(x)$ and $N_k(x)$ are polynomials in $x$ with integer coefficients and of degree less than or equal to $k$ for all nonnegative integer $k$.
How do I show this? Based on what I have seen online, I have to use mathematical induction and show it holds for $k+1$ i.e.,
\begin{equation*}
    B_{k+1}(x) = (k+1)!(M_{k+1}(x)\sin x + N_{k+1}(x)\cos x)
\end{equation*}

Notation: $\Bbb Z[x]$ is a standard notation for the set of polynomials in the free variable $x,$ with integer co-efficients.
Let $S(k)$ be the statement $T(k)\land T(k+1)$ where $T(k)$ is the statement that $B_k(x)=k![M_k\sin x +N_k\cos x]$ where $M_k, N_k\in \Bbb Z[x]$ with $\max (\deg M_k, \deg N_k)\le k.$
You have $S(0).$ Now prove that $\forall k\in \Bbb N_0\,[S(k)\implies S(k+1)].$
Note that $S(k)\implies  T(k+1)$ while  $[S(k+1)]\iff [T(k+1)\land T(k+2)] .$ So to prove $S(k)\implies S(k+1),$ it suffices to prove that $S(k)\implies T(k+2).$
