Computing $\int^{\pi/2}_0{\frac{1}{4\cos{x}+5}}$, $x \in \mathbb{R}$ by substituting $\cos{x} = (e^{ix}+e^{-ix})/2$?

Question

If an integral involves real trigonometric function, is substituting the definition via the complex exponential function a valid way of solving it?
Here is my example, I get the right answer but I'm not sure whether the derivation is valid, especially the part with complex logarithm: $$\int^{\frac{\pi}{2}}_0{\frac{1}{4\cos{x} + 5}}dx$$ $$\int{\frac{1}{4\cos{x} + 5}}dx = \int{\frac{1}{4\left(\frac{e^{ix}+e^{-ix}}{2}\right) + 5}}dx = \frac{1}{2}\int{\frac{1}{e^{ix}+e^{-ix} + \frac{5}{2}}}dx$$ Let $u = e^{ix}$, $du = ie^{ix}$, $dx = \frac{1}{iu}du$ $$\frac{1}{2}\int{\frac{1}{iu(u+u^{-1}+\frac{5}{2})}}du = \frac{1}{2i}\int{\frac{1}{u^2 + \frac{5}{2}u + 1}}du = \frac{1}{2i}\int{\frac{1}{(u+\frac{1}{2})(u+2)}}du$$ After PFD I get $$\frac{1}{3i}\int{\frac{1}{u+\frac{1}{2}}}du - \frac{1}{3i}\int{\frac{1}{u+2}}du = \frac{1}{3i}\left(\ln{(u+\frac{1}{2})} - \ln{(u+2)}\right) = \frac{1}{3i}ln{\left(\frac{e^{ix} + \frac{1}{2}}{e^{ix} + 2}\right)} =: F(x)$$
$$F(\frac{\pi}{2}) - F(0) = \frac{1}{3i}\left(\ln{\frac{e^{i\frac{\pi}{2}} + \frac{1}{2}}{e^{i\frac{\pi}{2}} + 2}} - \ln{\frac{e^0 + \frac{1}{2}}{e^0 + 2}}\right) = \frac{1}{3i}\left(\ln{\frac{\frac{1}{2} + i}{2 + i}} - \ln{\frac{1}{2}}\right) = \frac{1}{3i}\left(\ln{( \frac{2}{5} + \frac{3}{10}i)} - \ln{\frac{1}{2}}\right) = \frac{1}{3i}\ln{\left( \frac{4}{5} + \frac{3}{5}i\right)} = \frac{1}{3i}\left( \ln{\sqrt{(\frac{4}{5})^2 + (\frac{3}{5})^2 }} + i\arctan{\frac{(\frac{3}{5})}{(\frac{4}{5})}}\right) = \frac{1}{3i}\left(\ln{(1) + i\arctan{(\frac{3}{4})}}\right) = \frac{1}{3}\arctan{(\frac{3}{4})}$$

Answer 1

A faster way is to exploit real methods.

$$ \int_{0}^{\pi/2}\frac{dx}{5+4\cos x}\stackrel{x=2z}{=}2\int_{0}^{\pi/4}\frac{dz}{1+8\cos^2 z}\stackrel{z=\arctan u}{=}2\int_{0}^{1}\frac{du}{9+u^2}=\color{red}{\frac{2}{3}\arctan\frac{1}{3}}.$$

Answer 2

This is exactly the Kepler angle substitution: $$\begin{align}\sin\psi&=\frac{\sqrt{1-e^2}\sin x}{1+e\cos x}\\ \cos\psi&=\frac{\cos x+e}{1+e\cos x}\\ d\psi&=\frac{\sqrt{1-e^2}dx}{1+e\cos x}\end{align}$$ For $0<e<1$. So $$\begin{align}\int_0^{\pi/2}\frac{dx}{5+4\cos x}&=\frac15\frac1{\sqrt{1-e^2}}\int_0^{\cos^{-1}e}d\psi=\frac{\cos^{-1}e}{5\sqrt{1-e^2}}\\ &=\frac{\cos^{-1}\left(\frac45\right)}{3}=\frac13\tan^{-1}\left(\frac34\right)=\frac23\tan^{-1}\left(\frac13\right)\end{align}$$ With $e=\frac45$.