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
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Replying to @InertialObservr
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
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Replying to @art_sobolev
To find the derivative of arctan you need inverse function theorem, or at the very least to prove the chain rule
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Replying to @InertialObservr
Sure, but the (proposed) alternative is all of integral calculus
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Replying to @InertialObservr
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.
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Replying to @art_sobolev
It just means “that function whose derivative is f(x)” it’s just notation
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Replying to @InertialObservr
Won't the change of variables require chain rule then?
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Replying to @InertialObservr
Yep, and x = tan(t) with its derivative is half of the chain rule needed for the inverse theorem. Justification of this is the second half. Once you switched to the "no integrals here, just derivatives" interpretation, you started deriving the inverse theorem in disguise.
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You don’t need chain rule for the derivative of tan. Quotient rule does the trick
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