How did the answer found the antiderivative of $\frac{a}{{a}^{2}+{\mathrm{cos}}^{2}x}$ to be $\frac{1}{\sqrt{1+{a}^{2}}}{\mathrm{tan}}^{-1}(\frac{a\mathrm{tan}x}{\sqrt{1+{a}^{2}}})$?

I have not done Laplce transform previously and so is this purely by Laplace transform? I could verify that this is indeed the anti-derivative but could we get around Laplace transform to find the antiderivative straightaway?

I have not done Laplce transform previously and so is this purely by Laplace transform? I could verify that this is indeed the anti-derivative but could we get around Laplace transform to find the antiderivative straightaway?