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

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.

Answer 1

One approach would be to express the integrand as $$\frac{1}{4}\left(\frac{1}{2-\sqrt{3}\cos x} +\frac{1}{2+\sqrt{3}\cos x} \right) $$ and then note that the integral over $[0,\pi]$ for both integrands are equal so that the original integral is equal to $$\frac{1}{2}\int_{0}^{\pi}\frac{dx}{2+\sqrt{3}\cos x} $$ Next we put a highly non-obvious substitution $$(2+\sqrt{3}\cos x) (2-\sqrt{3}\cos y) = 1$$ to reduce the integral to $$\frac{1}{2}\int_{0}^{\pi}dy=\frac{\pi}{2}$$ More generally if $a>|b|$ then the substitution $$(a+b\cos x) (a-b\cos y) =a^{2}-b^{2}$$ yields the equations $$\sin x=\frac{\sqrt{a^{2}-b^{2}}\sin y} {a-b\cos y}, \, \frac{dx} {a+b\cos x} = \frac{dy} {\sqrt{a^{2}-b^{2}}}$$ so that $$\int_{0}^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt{a^{2}-b^{2}}}$$ Once again the above substitution comes straight from problem no $4$, page $266$ of Hardy's A Course of Pure Mathematics, 10th edition.

Answer 2

To avoid improper integrals for the half-angle substitution

$$\int_0^\pi \frac{1}{4-3\cos^2 x}\, dx = \int_0^{\pi/2} \frac{1}{4-3\cos^2 x}\, dx +\int_{0}^{\pi/2} \frac{1}{4-3\sin^2 x}\, dx $$

Then evaluate each one using $t = \tan(x/2)$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}