# Doubt on proper time explicit integration

I have a doubt on explicit calculation of proper time.

Considering that the metric is given by:

$$ds^{2} = -Adt^{2} + B^{-1}dr^{2}+Cd\Omega^{2} -2Ddtd\phi \tag{1}$$

where $$d\Omega^{2}$$ is the solid angle line element. Given that $$r=cte$$ , $$a=cte$$ , $$\theta = cte$$, then we can say that the proper time (infinitesimal) interval is given by:

$$d\tau ^{2} = g_{00}dt^{2} \equiv -Adt^{2} \tag{2}$$

I tried to calculate explicitly and I reached on:

$$\int_{0}^{\tau}d\tau = \int_{0}^{t} dt \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \tag{3}$$

My doubt is:

Since $$r=cte$$ , $$a=cte$$ , $$\theta = cte$$ (therefore the hole square root is a fixed number $$[*]$$) then, can I say that the proper time is given by the follwing expression?

$$\mathcal{T} = \int_{0}^{t} dt \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} =$$

$$= \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \int_{0}^{t} dt =$$

$$= \sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \hspace{3mm} t \implies$$

$$\mathcal{T} = \sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \hspace{3mm} t$$

$$* * *$$

$$[*]$$ In the software Mathematica, you would think about the $$r$$, $$\theta$$, and $$a$$, as quantities to insert on Manipulate structure, as just scrolling values. The real variable is therefore the time $$t$$.