Is there a simple way to show this?

Question

Prove without using the fundamental theorem of calculus:
∫ sec²(x) dx = tan(x) + C

Answer 1

I don't know if this is what you want, but we could start with the knowledge that
d(sec x) = secx tanx dx

Then ∫ sec^2 (x) dx = ∫ secx / tanx * secx tanx dx
= ∫ secx / √(sec^2 x - 1) * d(sec x)
= √(sec^2 x - 1)
= tan x

But now this raises the same question about
∫ secx tanx dx = sec x + C

------------------------
**COMPLEX VARIABLE METHOD**
Here's another method that does not require the knowledge of standard trig integrals.

Let's try to find ∫{0,θ} sec^2 θ dθ
Consider the unit circle in the complex plane, where
z = exp(i*θ) = x + iy
1/z = x - iy
dz = i*exp(i*θ) dθ = iz dθ
Also 2x = z + 1/z

On the unit circle, secθ = 1/x
sec^θ = 1/x^2 = 4z^2 / (z^2 + 1)^2
so we desire to integrate 1/x^2 on a segment of the unit circle from 0 to θ (<π/2).
The reason we wish to avoid θ = π/2 is because
4z^2 / (z^2 + 1)^2 has a pole at ±i.
Also we know that secθ and tanθ are both undefined at θ = π/2. So let's restrict this segment to the first quadrant.

The integral transforms to:
-2i* ∫{1,z} 2z dz / (z^2 + 1)^2

***EDIT*** in the above line, I initially had the integral going from 0 to z, which is why I was getting an extra term in the final result. It should go from 1 to z, since on the unit circle at θ = 0, z = 1.

= 2i * [1/(z^2 + 1) - 1/2]

Substituting z = exp(iθ) and simplifying we get
2i*[1 / (1 + cos2θ - i*sin2θ) - 1/2]
= 2i* (-1/2 i tanθ)
= tanθ

*EDIT*
I figured out the earlier error. See above.
Also with this method, you can determine the antiderivatives of quite a number of trig functions without prior knowledge of other standard trig integrals.

Answer 2

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Answer 3

sec(x)=hyp/adj. int(hyp^2/adj^2)= int((adj^2+opp^2)/adj^2)= int(adj^2/adj^2+opp^2/adj/^2)= int(1+tan^2(x))= int(1+u^2)=int(1)+int(u^2)= x+u^3/3*(sec^2(x))= x+tan^3(x)/(sec^2(x))...

Can't quite get there, but I'd imagine that's the way to start.