calculus - $n^2 int_0^1 (1-x)^n sin(pi x) mathrm{d}x$

I would like to find :

$$\lim_{n \to \infty} n^2 \int_0^1 (1-x)^n \sin(\pi x) \mathrm{d}x $$

We have :
$$n^2 \int_0^1 (1-x)^n \sin(\pi x) \mathrm{d}x = \int_0^1 n^2(1-x)^n \sin(\pi x) \mathrm{d}x$$

Moreover we have $\forall x \in [0, 1]$ :

$$\lim_{n \to \infty} n^2(1-x)^n \sin(\pi x) = 0$$

So by the dominated convergence theorem we can deduce that :

$$ \lim_{n \to \infty} \int_0^1 n^2(1-x)^n \sin(\pi x) \mathrm{d}x = 0$$

Yet, here my book say the answer is actually $\pi$, and I don't understand why what I've done is wrong, and how I can actually find that the value is $\pi$.

Answer

You cannot apply the DCT because there is not an integrable function $g$ (independent of $n$) such that $n^2(1-x)^n\sin(\pi\,x)\le g(x)$.

Integrating by parts twice we get
$$
\int_0^1(1-x)^n\sin(\pi\,x)\,dx=\frac{\pi}{(n+1)(n+2)}+\frac{\pi^2}{(n+1)(n+2)}\int_0^1(1-x)^{n+2}\sin(\pi\,x)\,dx.
$$
Multiply by $n^2$, let $n\to\infty$ and observe that the DCT can be applied to the integral on the left hand side because $n^2/(n+1)/(n+2)$ is bounded.