〈 Berger | Dillon 〉

Isn't it more straightforward to derive the same through atan's derivative? atan(x) = ∫ˣ₀ 1/(1+t²) dt

Oh how great our vision is in hindsight! In a way, yes.. If you’re clever enough to come up with trig sub before it was invented.. or you knew the derivative or arctan(x) The beauty of this trick is that it doesn’t require anything except knowledge of derivatives

Well, the derivative of arctan is easy to come up with if you know tan's derivative and start with the identity f⁻¹(f(x)) = x The problem I have with your trick is that 1/cos²(t) appeared out of nowhere, I don't see what it teaches us and how to apply it elsewhere

To find the derivative of arctan you need inverse function theorem, or at the very least to prove the chain rule

Sure, but the (proposed) alternative is all of integral calculus

How so? No integration is done

But what does this stretched out S even mean then? What is t and dt, and how come it suddenly got replaced by x and dx? Besides, that's not true, you integrated 1 in the very first line.