Clarification of proof that $\pi ^2$ is irrational

Question

My textbook Elementary analysis and the theory of calculus has a proof that $\pi ^2$ is irrational. I will skip some steps for simplification.

The proof starts with a claim that there exists a sequence of polynomials with integer coefficients that $(Q_n)_{n=0}^{\infty}$ = $I_n$ where $I_n$ the sequence of integrals $I_n = \int_0^{\pi}P_n(x)\sin x\, dx$ where $P_n(x) = \frac {(x(\pi - x))^n}{n!}$ . First the author shows with integration by parts that

$$(1)\qquad I_n = -\int_0^{\pi}P''_n(x)\sin x\, dx.$$

This is fine with me. Next he shows that $P'_n(x)$ = $P_{n-1}(x)(\pi - 2x)$ and uses this to show that for $n \geq 2$ that

$$(2)\qquad P''_n(x) = \pi^2P_{n-2}(x) - (4n-2)P_{n-1}(x).$$

This is fine as well but where I get confused is the next statement which claims that because of $(1)$ and $(2)$, $I_n = -\pi^2 I_{n-2} + (4n-2)I_{n-1}$.

I suspect this uses the sum rule for integrals and the substitution in $(1)$ but any clarification would be much appreciated. I can also add in more intermediate steps I have skipped if necessary.

Answer 1

Note that

\begin{align*} I_n &= -\int_0^{\pi}P_n''(x)\sin x\, dx\\ &= -\int_0^{\pi}(\pi^2P_{n-2}(x) - (4n-2)P_{n-1}(x))\sin x\, dx\\ &= -\pi^2\int_0^{\pi}P_{n-2}(x)\sin x\, dx + (4n - 2)\int_0^{\pi}P_{n-1}(x)\sin x\, dx\\ &= -\pi^2 I_{n-2} + (4n-2)I_{n-1}. \end{align*}