How to calculate $\int_{0}^1 \frac{(\cos x-1)^2}{x^2} $?

How to calculate $\int_{0}^1 \frac{(\cos x-1)^2}{x^2} $ ? I tried integration by parts but it only got worse.(I don't want to use Si$(x)$ function) Any ideas?

Expand the numerator: $(1-\cos x)^2 = 1 -2\cos x + \cos^2x = 3/2 -2\cos x + \cos(2x)/2$, and notice that $$ \int \frac{\cos(ax)}{x^2} \mathrm{d}x = -a\mathrm{Si}(ax) - \frac{\cos(ax)}{x} + C, $$ where $\mathrm{Si}(x)$ is the sine integral. You can solve it integrating by parts.

This method will give you as a result: $$ \int \frac{(1-\cos x)^2}{x^2} \mathrm{d}x = 2\mathrm{Si}(x) - \mathrm{Si}(2x) - \frac{4\sin^4\left(\frac{x}{2}\right)}{x} + C. $$

Therefore, $$ \boxed{\int_0^1 \frac{(1-\cos x)^2}{x^2} \mathrm{d}x = 2\mathrm{Si}(1) - \mathrm{Si}(2) - 4\sin^4\left(\frac{1}{2}\right) \approx 0.0754312.} $$

If you do not want to use the sine integral function, integrate a series representation of the integrand, for example, but I do not understand why you do not want to use it if it was created explicitly for this cases.