Step 1: Identify the condition on $a$.
$a > |\sec\theta| \geq 1$, so $a > 1$. This means $a - 1 > 0$ and $a + 1 > 0$, so $\sqrt{a-1}$ and $\sqrt{a+1}$ are both real and positive.
Step 2: Use Weierstrass substitution $t = \tan(x/2)$.
Then $\cos x = \dfrac{1-t^2}{1+t^2}$ and $dx = \dfrac{2\,dt}{1+t^2}$.
Step 3: Substitute into the integral.
\[ I = \int \frac{1}{1 + a\cdot\frac{1-t^2}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = \int \frac{2\,dt}{(1+t^2) + a(1-t^2)} = \int \frac{2\,dt}{(1+a) + (1-a)t^2}. \]
Step 4: Rewrite the denominator.
Since $a > 1$, $1 - a < 0$. So denominator $= (1+a) - (a-1)t^2$. \[ I = \int \frac{2\,dt}{(1+a)\left(1 - \frac{a-1}{a+1}t^2\right)} = \frac{2}{a+1}\int \frac{dt}{1 - \frac{a-1}{a+1}t^2}. \]
Step 5: Use $\int \frac{dt}{1 - k^2 t^2} = \frac{1}{2k}\ln\left|\frac{1+kt}{1-kt}\right|$.
Here $k^2 = \dfrac{a-1}{a+1}$, so $k = \sqrt{\frac{a-1}{a+1}}$. The integral becomes: \[ \frac{2}{a+1} \cdot \frac{1}{2k} \ln\left|\frac{1 + kt}{1 - kt}\right| = \frac{1}{(a+1)k} \ln\left|\frac{1+kt}{1-kt}\right|. \]
Step 6: Substitute back and simplify.
$(a+1)k = (a+1)\sqrt{\frac{a-1}{a+1}} = \sqrt{(a+1)(a-1)} = \sqrt{a^2-1}$. With $t = \tan(x/2)$ and $kt = \sqrt{\frac{a-1}{a+1}}\tan\frac{x}{2} = \frac{\sqrt{a-1}\sin(x/2)}{\sqrt{a+1}\cos(x/2)}$: \[ I = \frac{1}{\sqrt{a^2-1}} \ln\left|\frac{\sqrt{a+1}\cos\frac{x}{2} + \sqrt{a-1}\sin\frac{x}{2}}{\sqrt{a+1}\cos\frac{x}{2} - \sqrt{a-1}\sin\frac{x}{2}}\right| + C. \] \[ \boxed{\dfrac{1}{\sqrt{a^2-1}}\ln\left(\dfrac{\sqrt{a+1}\cos\frac{x}{2}+\sqrt{a-1}\sin\frac{x}{2}}{\sqrt{a+1}\cos\frac{x}{2}-\sqrt{a-1}\sin\frac{x}{2}}\right) + C} \]