Alternate Derivation of SHM Equation

here is a derivation of the SHM equation in a slightly simpler form that might be easier to understand for those who don't want to go through all linear operations and auxiliary equations that comes with the other proof.
so here is a different proof where i've just used a few substitutions and a couple integrations.

we all know the basic condition for SHM:
F = -kx
and Newton's Second Law:
F = ma

and from these we construct our differential equation (look to the previous blog post if you need to see how this is put together):
d²c/dt² + c²x = 0,
where c² = k/m

now, let v = dx/dt. then also, d²x/dt² = v * dv/dx (this is a substitution often used for solving differential equations)
v * dv/dx = -c² x
v dv = -c² x dx
∫v dv = -c² ∫x dx + C₁
½ v² = -c² ½ x² + C₁
v² = -c² x² + C₂ ... (C₂ = 2C₁)
(dx/dt)² = -c² x² + C₂

let us now find C₂:
we assume that at time t = 0, the oscillating object is at its maximum displacement, A, and has no velocity. no velocity means that x(0) = A, x'(0) = 0:
x'(0) = -c² x(0)² + C₂
0² = -c² A² + C₂
C₂ = c²A²

now back to the differential equation:
(dx/dt)² = -c²x² + c²A²
dx/dt = ±√[-c²x² + c²A²]
dx/dt = ± c√[ A² - x²]
dx / √[ A² - x²] = ± c dt
∫dx / √[ A² - x²] = ± c ∫dt + C₃

here we use a trigonometric substitution x = A sinθ to solve the integral. this makes:
√[ A² - x² ] = √[ A² - A² sin²θ ] = A √ (1 - sin²θ) = A √ (cos²θ) = A cosθ
and,
dx = A cosθ dθ
lastly,
sinθ = x/A => θ = arcsin[x/A]

∫A cosθ dθ/ A cosθ = ± c ∫dt + C₃
∫dθ = ± c ∫dt + C₃
θ = ± ct + C₃
arcsin[x/A] = ± ct + C₃
x/A = sin( ±ct + C₃ )
x = A sin( ±ct + C₃ )

now, to find C₃:
recall that at time t = 0, x is a maximum of A:
x(0) = A sin( ±c*0 + C₃)
A = A sin(C₃)
1 = sin(C₃)
C₃ = π/2

then our equation becomes:
x = A sin( ±ct + π/2 ), but know from trigonometry that sin(θ + π/2) = cos(θ)
x = A cos( ±ct)
and remember also that cos(-θ) = cos(θ), which gives:
x(t) = A cos(ct)

THERE WE GO!
see, that wasn't so difficult! it was actually probably easier to follow than the previous derivation.
and i wrote out literally ALL the steps necessary. a physics textbook wouldn't have to show half of this stuff, and it's not that difficult to put in the back in an appendix or something, come on!