TITLE: Is there a change of variables that allows one to calculate $\int_0^\pi \frac{1}{4-3\cos^2 x}\, \mathrm dx$ avoiding improperties?
QUESTION [5 upvotes]: I'm trying to evaluate this integral: 
$$\int_0^\pi \frac{1}{4-3\cos^2 x}\, \mathrm dx$$
It is obvious that this is a standard Riemann integral (without improperties)
BUT
The classical $\tan x=t$ changes of variables introduces 2 improperties: ($x=(\pi/2)^-$ and $x=(\pi/2)^+$).
The other possibility is $\tan(x/2)=t$, but this change introduce an improperty in $x=\pi$.
So, my question is:
Is there any change of variables (or integral method) that allows one to calculate this integral avoiding improperties? That is, only using standard Riemann integrals.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{\pi}{\dd x \over 4 - 3\cos^{2}\pars{x}} & =
\int_{0}^{\pi/2}{\dd x \over 4 - 3\cos^{2}\pars{x}} +
\int_{\pi/2}^{\pi}{\dd x \over 4 - 3\cos^{2}\pars{x}}
\\[5mm] &=
2\int_{0}^{\pi/2}{\dd x \over 4 - 3\cos^{2}\pars{x}} =
2\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over 4\sec^{2}\pars{x} - 3}
\\[5mm] & =
\int_{0}^{\pi/2}{2\sec^{2}\pars{x}\,\dd x \over 4\tan^{2}\pars{x} + 1}
\\[5mm] & \stackrel{t\ \equiv\ 2\tan\pars{x}}{=}\,\,\,
\int_{0}^{\infty}{\dd t \over t^{2} + 1} = \bbx{\ds{\pi \over 2}}
\end{align}