(35S) Quick integration

[Albert Chan]

(11-28-2022 06:49 PM)Albert Chan Wrote:  At the end of blackpenredpen video, we have:

∫(√(x^2+1) dx) = (x*√(x^2+1) + ln(x+√(x^2+1))) / 2 + C

Another way, by guessing shape of RHS

Let x = sinh(y)      → dx = cosh(y) dy

Matching integral, we wanted cosh(y) dx
Guess from integration by part term, cosh(y)*x

d(cosh(y)*x) = d(sinh(2y)/2) = cosh(2y) dy = (2*cosh(y)^2-1) dy = 2*cosh(y) dx - dy

cosh(y) dx = (d(x*√(x^2+1)) + dy) / 2

--> ∫(√(x^2+1) dx) = (x*√(x^2+1) + asinh(x)) / 2 + C

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Mathematica way, let x = tan(y), dx = sec(y)^2 dy

∫(√(x^2+1) dx) = ∫(sec(y)^3 dy)

∫(sec(y)^3 dy)
= ∫(sec(y) d(tan(y))
= tan(y)*sec(y) - ∫(tan(y) * (tan(y) sec(y) dy))      // integrate by parts
= tan(y)*sec(y) + ∫(sec(y) dy) - ∫(sec(y)^3 dy)

From definition of Gudermannian function inverse


∫(sec(y)^3 dy) = (tan(y)*sec(y) + asinh(tan(y))) / 2 + C

--> ∫(√(x^2+1) dx) = (x*√(x^2+1) + asinh(x)) / 2 + C

This is also the approach used by https://www.integral-calculator.com/