Let's say that we want to evaluate ∫0∞1+x21dx without knowing the anti derivative of 1+x21. How could we do this? To start let's consider the unit circle, we know the unit circle has arc-length L=rθ=θ. We can equivalently write the arc length of the unit circle as
L=====∫0θdt∫0θcos2(t)1cos2(t)1dt∫0θcos2(t)sin2(t)+cos2(t)cos2(t)1dt∫0θ1+tan2(t)cos2(t)1dt∫0tan(θ)1+x21dx via x=tan(t).
Notice when θ→2π, tan(θ)→∞. So recalling that L=θ:
θ=2π=∫0tan(θ)1+x21dx∫0∞1+x21dx.
We can also use this to discover many other relations. For example
4π=3π=−2π=∫011+x21dx∫031+x21dx∫0−∞1+x21dx.
This was the method of Italian mathematician Giulio Fangnano.
We can of course immediately infer via the inverse function of tan(θ) the anti-derivative in question. We know via the fundamental theorem of calculus
θ=θ=∫0tan(θ)1+x21dx=F(tan(θ))−F(0),F(0)=0F(tan(θ)).
And we know this is in the form f−1(f(x))=x. So we can infer F(x)=arctan(x)+C, which leads us to the anti-derivative we assumed we didn't know at the beginning
∫1+x21dx=arctan(x)+C.
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