# How did the answer found the antiderivative of (a)/(a^2+cos^2 x) to be (1)/(sqrt(1+a^2)) tan^(-1) ((a tan x)/(sqrt(1+a^2))) ?

How did the answer found the antiderivative of $\frac{a}{{a}^{2}+{\mathrm{cos}}^{2}x}$ to be $\frac{1}{\sqrt{1+{a}^{2}}}{\mathrm{tan}}^{-1}\left(\frac{a\mathrm{tan}x}{\sqrt{1+{a}^{2}}}\right)$?
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?
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Waylon Jenkins
When the integrand has only even powers of $\mathrm{sin}x$ and $\mathrm{cos}x$, the substitution $z=\mathrm{tan}x$ can be useful:
$\begin{array}{rl}\int \frac{\alpha }{{\alpha }^{2}+{\mathrm{cos}}^{2}x}dx& =\int \frac{\alpha dz}{{\alpha }^{2}+1+{\alpha }^{2}{z}^{2}}=\int \frac{d\theta }{\sqrt{1+{\alpha }^{2}}}\\ & =\frac{\theta }{\sqrt{1+{\alpha }^{2}}}+C=\frac{1}{\sqrt{1+{\alpha }^{2}}}{\mathrm{tan}}^{-1}\frac{\alpha z}{\sqrt{1+{\alpha }^{2}}}+C\\ & =\frac{1}{\sqrt{1+{\alpha }^{2}}}{\mathrm{tan}}^{-1}\frac{\alpha \mathrm{tan}x}{\sqrt{1+{\alpha }^{2}}}+C\end{array}$
Where we have made the further substitution $\alpha z=\sqrt{1+{\alpha }^{2}}\mathrm{tan}\theta$

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