options - Black-Scholes formula proof, without stochastic integration

I've looked into many books at my academic library, and very often it goes like this:

• Brownian motion



• Then, stochastic integration (Itô's formula etc.)


• Application: Black-Scholes formula for price of a call option

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However, I've seen a proof that doesn't require stochastic integration at all, it goes like this:

Let $Y(t) = Y(0) e^{X(t)}$ be the price of an asset, as a geometric brownian motion, i.e. $X(t) = \mu t + \sigma B(t)$ where $B$ is a standard brownian. We also assume that $\mu + \sigma^2 / 2 = 0$ so that the trend is neutral.

Now the price for a European call option (maturity $t=T$, strike price $Y(0) A$) is:

$$C = E\big( (Y(T)-Y(0) A)^+ \big) = Y(0) \cdot E\big( (e^{X(T)}-A)^+ \big).$$

But since $X(T)$ has a law $\mathcal{N}(\mu T, \sigma^2 T)$ (brownian), it's easy to see that

$$E\big( (e^{X(T)}-A)^+ \big) = \int_{\ln A}^\infty (e^x-A)\frac{1}{\sqrt{2\pi \sigma^2 T}} e^{-\frac{(x-\mu T)^2}{2 \sigma^2 T} } d x$$

and then (it's just standard integration), we get:

$$C = Y(0) (\Phi(\sigma \sqrt{T} - \alpha_T) - A \Phi(-\alpha_T))$$ with $\alpha_T := \ln(A)/(\sigma\sqrt T)+\sigma\sqrt T/2$ and $\Phi(x) = (1/\sqrt{2 \pi}) \int_{-\infty}^x e^{-t^2/2} d t$

This proves the Black-Scholes formula for a call option, without needing any stochastic integration / Itô.

Why do all textbooks use stochastic integration to prove this, when it seems we don't need it?