Wherever I look on the internet it says that sinx/x cannot be integrated by elementary techniques, but it seems that there is a method using integration by parts of the quotient rule. However, when I compute definite integrals with this, the answer that my calculator returns is different than the answer I get with the definite integral. Can anyone tell me where my problem is or if my method does actually work.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}[/tex]

[tex]\frac{f(x)}{g(x)}=\int \frac{f'(x)g(x)}{g^2(x)}\,dx-\int \frac{g'(x)f(x)}{g^2(x)}\,dx[/tex]

[tex]\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx[/tex]

[tex]\int \frac{sin(x)}{x}\,dx=\int \frac{xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx[/tex]

[tex]g(x)=x^2[/tex]

[tex]f(x)=sin(x)[/tex]

[tex]g'(x)=2x[/tex]

[tex]f'(x)=cos(x)[/tex]

[tex]\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx[/tex]

[tex]\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{cos(x)}{x}\,dx-\frac{sin(x)}{x}\,dx[/tex]

[tex]\int \frac{cos(x)}{x}\,dx=\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx[/tex]

[tex]\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx[/tex]

[tex]\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx[/tex]

[tex]\int \frac{sin(x)}{x}\,dx=\frac{1}{2}\left(\frac{1}{2} \int \frac{-sin(x)}{x} \,dx - \frac{1}{2}\frac{cos(x)}{x}\right)-\frac{1}{2}\frac{sin(x)}{x}[/tex]

[tex]\frac{5}{4}\int \frac{sin(x)}{x}\,dx=-\frac{1}{4}\frac{cos(x)}{x}-\frac{1}{2}\frac{sin(x)}{x}[/tex]

[tex]\int \frac{sin(x)}{x}\,dx=-\frac{1}{5}\frac{cos(x)}{x}-\frac{2}{5}\frac{sin(x)}{x}[/tex]

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# Integration of sinx/x

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