Let π=a/b, the quotient of positive integers. We define the polynomials
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% aaaa!6D57!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$\begin{gathered}
f(x) = \frac{{{x^n}{{(a - bx)}^n}}}{{n!}}, \hfill \\
F(x) = f(x) - {f^{(2)}}(x) + {f^{(4)}}(x) - \cdot \cdot \cdot + {( - 1)^n}{f^{(2n)}}(x), \hfill \\
\end{gathered}$$</EquationSource></Equation>
the positive integer <Emphasis Type="Italic">n</Emphasis> being specified later. Since <Emphasis Type="Italic">n!f(x)</Emphasis> has integral coefficients and terms in <Emphasis Type="Italic">x</Emphasis> of degree not less than <Emphasis Type="Italic">n</Emphasis>, <Emphasis Type="Italic">f(x)</Emphasis> and its derivatives <Emphasis Type="Italic">f<Superscript>(<Emphasis Type="Italic">i</Emphasis>)</Superscript>(x)</Emphasis> have integral values for x=0; also for <Emphasis Type="Italic">x</Emphasis>=π<Emphasis Type="Italic">=a/b</Emphasis>, since <Emphasis Type="Italic">f(x) =f(a/b−x)</Emphasis>. By elementary calculus we have <Equation ID="Equb"><EquationSource Format="MATHTYPE"><![CDATA[% MathType!MTEF!2!1!+-
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]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$\frac{d}{{dx}}\left\{ {F'(x)\sin x - F(x)\cos x} \right\} = F''(x)\sin x + F(x)\sin x = f(x)\sin x$$
and
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% Gaey4kaSIaamOraiaacIcacaaIWaGaaiykaaaa!60A7!]]</EquationSource><EquationSource Format="TEX"><![CDATA[$$\int_0^\pi {f(x)\sin xdx = \left[ {F'(x)\sin x - F(x)\cos x} \right]} _0^\pi = F(x) + F(0)$$
(1)
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