I saw this question earlier. I was wondering how did the answer found the antiderivative of $\frac{a}{a^2+\cos^2x}$ to be $\frac{1}{\sqrt{1+a^2}}\tan^{-1}(\frac{a \tan x}{\sqrt{1+a^2}})$?
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?
When the integrand has only even powers of $\sin x$ and $\cos x$, the substitution $z=\tan x$ can be useful: $$\begin{align}\int\frac{\alpha}{\alpha^2+\cos^2x}dx&=\int\frac{\alpha dz}{\alpha^2+1+\alpha^2z^2}=\int\frac{d\theta}{\sqrt{1+\alpha^2}}\\ &=\frac{\theta}{\sqrt{1+\alpha^2}}+C=\frac1{\sqrt{1+\alpha^2}}\tan^{-1}\frac{\alpha z}{\sqrt{1+\alpha^2}}+C\\ &=\frac1{\sqrt{1+\alpha^2}}\tan^{-1}\frac{\alpha\tan x}{\sqrt{1+\alpha^2}}+C\end{align}$$ Where we have made the further substitution $\alpha z=\sqrt{1+\alpha^2}\tan\theta$
