Is this proof about $\pi$ is irrational correct?

A proof is

If $\pi$ is rational, then $\pi$=$\frac{a}{b}$ where $a,b \in \mathbb{N}$

let $f(x)=x^n\left[\dfrac{(a-bx)^n}{n!}\right]$ , for $0<x<\frac{a}{b}$

$0<f(x)<\dfrac {\pi^na^n} {n!}$ , $0<\sin x<1$

thus $0<f(x)\sin x<\dfrac {\pi^na^n} {n!}$

where $n$ is big enough

$0<\int _{0}^{\pi}f(x)\sin x\mbox{d}x<1$ $\left( 1\right)$

let $F(x)=\sum _{i=0}^{n}(-1)^if^{(2i)}\left ( x\right)$ ($f^{(i)}\left( x\right)$ means $i$th derivative )

because $n!f(x)$ is a polynomial with integer coefficient, and all the degrees is bigger than n, so $f(x)$ and its i th derivatives at $x-0$ are all integer thus $F(0)$ and $F(\pi)$ are also integers

$\dfrac{\mbox{d}}{\mbox{d}x}[F'(x)\sin x-F(x)\cos x]=F''(x)\sin x+F'(x)\cos x-F'(x)\cos x+F(x)\sin x=F''(x)\sin x+F(x)\sin x=f(x)\sin x$

thus $\int _{0}^{\pi}f(x)\sin x\mbox{d}x=\int_{0}^{\pi}[F'(x)\sin x-F(x)\cos x]\mbox{d}x==F(π)+F(0)$

so $\int _{0}^{\pi}f(x)\sin x\mbox{d}x$ is integer contradicts with $(1)$.

I think this proof has some error but I cannot find it.

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This is Niven's proof. The integral (1) should be from $0$ to $\pi$. Also, the word is "contradicts" not "contracted". – user02138 Jul 24 '13 at 17:22
– dtldarek Jul 24 '13 at 17:30
Why $f=F+F''$? ${}$? – Kunnysan Jul 24 '13 at 17:30
We generally write $f^{(i)}$ for the $i$th derivative, just to distinguish between exponents. – Thomas Andrews Jul 24 '13 at 17:34