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## Recipe for a Leibniz Quarter Pi

Ingredient: $\frac{1}{1+x^2}$

Divide gently in a long division sauce pan:

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+x^8-x^{10}...$

Integrate briskly over a low flame:

$\int (1-x^2+x^4-x^6+x^8-x^{10}...)dx=$

$=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

Evaluate for $x=1$ and let stand at room temperature for 1000 terms for accuracy to three decimal places.

$=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$

But this doesn’t look like a pi! or taste like a pi!

or a quarter pi!

$\int \frac{1}{1+x^2}dx$

and make the trigonometric substitution $x=tan \theta$ so that $dx=sec^2 \theta d \theta$ then:

$\int \frac{1}{1+x^2}dx=\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta$

and

$\int \frac{1}{1+tan^2 \theta}sec^2 \theta d \theta=\int \frac{1}{sec^2 \theta}sec^2 \theta d \theta$

$=\int d \theta=\theta$

which, if we return to the original substitution $x=tan \theta$, we see that $\tan^{-1} x=\theta$

So, $\int \frac{1}{1+x^2}dx=\tan^{-1}x$, which means that:

$\tan^{-1}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+...$

and since $\tan^{-1}(1)=\frac{\pi}{4}$, then

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}...$