*inverse*trigonometric functions, which are hard for me to think about, but let's go slowly and hope for the best:

x is the angle whose sin is y. We draw a picture, and see that:

and

since we're dealing with the inverse function

`sin`^{-1} y

:slope of inverse = 1 / slope of original function

Since

`sin`^{-1} y + cos^{-1} y = π/2

, if:And our goal, remembering from here:

So, if we integrate

`1/(1 + y`^{2}) dy

, we get `tan`^{-1} y

. We're going to use that.The geometric series is:

How do we derive this? One way (not legal) is to assume that the series really does converge to a sum S

Or we can check it by just multiplying out:

Replace x by -x

^{2}

If we integrate both sides, the rhs is

`tan-1 x`

. And the lhs is:The magic: if we let

`x = 1`

, then:Another series for π, how cool is that?