Proving $\pi$ is irrational with recursive integrals Given $$I_n = \int_{-1}^1 (1-x^2)^n\cos(\theta x)dx,$$
I have proven the relation
$$\theta^2I_n = 2n(2n-1)I_{n-1} - 4n(n-1))I_{n-2}.$$
I am supposed to now notice that
$$\theta^{2n+1}I_n = n!(P_n(\theta)\sin(\theta) + Q_n(\theta)\cos(\theta)),$$
where $P_n$ and $Q_n$ are polynomials of degree at most $2n$ with integer coefficients, then use this relation to prove that $\pi$ is irrational. I can sort of see the relation, but can't see a clear way to prove it besides ''the coefficients seem to decrease by one each time so the $n!$ makes sense, and $I_1,I_0$ have the desired $\sin\theta$ and $\cos\theta$, and the plus $1$ makes sense since we apply the relation either $n$ or $n-1$ times on the terms (giving us $\theta^{2n}$ or $\theta^{2n-2}$) but then $I_1 = \frac{4}{\theta^3}(\sin\theta - \theta\cos\theta)$ and $I_0 = \frac{2}{\theta}\sin\theta$''. So, I can vaguely see how all the parts could fit together but am not convinced with my argument and don't see an obvious way to prove it carefully by actually applying the relation by hand (I've tried expanding a few of the term by hand to see if a particularly nice pattern appears, but I don't see one). Can someone explain how to do this a bit more carefully? I see how to prove that $\pi$ is irrational given this relation (plug in $\theta = \pi/2$ and find an integer between $0$ and $1$ assuming $\pi$ is rational).
I got the problem from this practice sheet: [https://www.dpmms.cam.ac.uk/study/IA/AnalysisI/2020-2021/aI_4_21.pdf]
 A: $$\theta^2 I_n = 2n(2n-1)I_{n-1} - 2n(2n-2)I_{n-2}\tag 1$$
You can do it directly by iteratively multiplying both sides of $(1)$ by $\theta^2$ and reducing it, but that seems tedious.

Looks like it will be easier to prove it using strong induction on $n$.
Show the base case holds. Assume the induction hypothesis for $k\le n$, ie,

$$\forall k\le n\colon\quad\theta^{2k+1}I_k = k!(P_k(\theta)\sin(\theta) + Q_k(\theta)\cos(\theta))\tag{$\dagger$}$$

Now, show the inductive step for $n+1$, ie,
$$\begin{align}\theta^{2(n+1)+1}I_{n+1}&=\theta^{2n+1}\theta^2 I_{n+1}\\&\overset{(1)}{=}\theta^{2n+1}\bigl((2n+2)(2n+1)I_n-(2n+2)(2n)I_{n-1}\bigr)\\&=(2n+2)(2n+1)[\theta^{2n+1}I_n]-(2n+2)(2n)\theta^2[\theta^{2(n-1)+1}I_{n-1}]\\&\overset{(\dagger)}{=}(2n+2)(2n+1)n!\bigl(P_n(\theta)\sin\theta+Q_n(\theta)\cos\theta\bigr)-\theta^2 (2n+2)(2n)\bigl((n-1)!P_{n-1}(\theta)\sin\theta+Q_{n-1}\cos\theta\bigr)\\&=2(2n+1)(n+1)!\bigl(P_n(\theta)\sin\theta+Q_n(\theta)\cos\theta\bigr)-4(n+1)!\bigl(\theta^2 P_{n-1}(\theta)\sin\theta+\theta^2 Q_{n-1}\cos\theta\bigr)\\&=(n+1)!(P_{n+1}(\theta)\sin\theta+Q_{k+1}(\theta)\cos\theta)\end{align}$$
where $P_{n+1}(\theta):=2(2n+1)P_n(\theta)-4\theta^2 P_{n-1}(\theta)$ and $Q_{n+1}(\theta):=2(2n+1)Q_n(\theta)-4\theta^2 Q_{n-1}(\theta)$
and we can finish noting that $P_{n+1}$ and $Q_{n+1}$ are both polynomials with integer coefficients and degree at most $2n+2=2(n+1)$
