Pruébalo, $\int_{0}^{2\pi}\frac{\cos x+2}{5+4\cos x} dx=\pi$

Question

He tratado de resolver esto que va como sigue-
$$\begin{align*} \int\frac{\cos x+2}{5+4\cos x} dx &=\int\frac{(1/4)(5+4\cos x)+(3/4)}{5+4\cos x} dx\\ &={1\over4}\int dx+{3\over4}\int\frac{dx}{9\cos^2 {x\over 2}+\sin^2{x\over2}}\\ &=({x\over4}+C)+{3\over4}\int\frac{\sec^2{x\over2}}{9+\tan^2{x\over2}}dx\\ &=({x\over4}+C)+{3\over2}\int\frac{dy}{9+y^2}\qquad \text{substituting $\tan {x\over2}=y$}\\ &={x\over4}+{1\over2}\arctan({y\over3})+C'={x\over4}+{1\over2}\arctan({\tan{x\over2}\over3})+C' \end{align*}$$ Así que, $$\int_{0}^{2\pi}\frac{\cos x+2}{5+4\cos x} dx=\left[{x\over4}\right]_{0}^{2\pi}+{1\over2}\left[\arctan({\tan{x\over2}\over3})\right]_{0}^{2\pi} \\={\pi\over2}+{1\over2}\left[\arctan({\tan\pi\over3})-0\right]={\pi\over2}+{1\over2}\left[0-0\right]={\pi\over2}$$ Así que, no puedo conseguir $\pi$ ¡¡!! ¿Puede alguien averiguar dónde está mi fallo? Gracias por la ayuda de antemano.

Answer 1

Tenga en cuenta que $$\int\frac{\cos x+2}{5+4\cos x} dx=F(x):={x\over4}+{1\over2}\arctan\left({\tan{x\over2}\over3}\right)+c$$ pero la función $F$ no es continua en el intervalo de integración $[0,2\pi]$ (por otro lado una primitiva de una función continua debe ser continua por la Teorema fundamental del cálculo ).

Desde $F$ tiene un "salto" en $x=\pi$ podemos dividir el intervalo y por lo tanto obtenemos $$\int_0^{2\pi}\frac{\cos x+2}{5+4\cos x} dx =\int_0^{\pi} +\int_{\pi}^{2\pi} =[F(x)]_{0^+}^{\pi^-}+[F(x)]_{\pi^+}^{2\pi^-}=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{2\pi}{\cos\pars{x} + 2 \over 5 + 4\cos\pars{x}}\,\dd x} = {\pi \over 2} + {3 \over 4}\int_{-\pi}^{\pi}{\dd x \over 5 - 4\cos\pars{x}} = {\pi \over 2} + {3 \over 2}\int_{0}^{\pi}{\dd x \over 5 - 4\cos\pars{x}} \\[5mm] = &\ {\pi \over 2} + {3 \over 2}\int_{-\pi/2}^{\pi/2}{\dd x \over 5 + 4\sin\pars{x}} = {\pi \over 2} + {3 \over 2}\int_{0}^{\pi/2}\bracks{{1 \over 5 + 4\sin\pars{x}} + {1 \over 5 - 4\sin\pars{x}}}\dd x \\[5mm] = &\ {\pi \over 2} + 15\int_{0}^{\pi/2}{\dd x \over 25 - 16\sin^{2}\pars{x}} = {\pi \over 2} + 15\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over 25\sec^{2}\pars{x} - 16\tan^{2}\pars{x}} \\[5mm] = &\ {\pi \over 2} + 15\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x \over 9\tan^{2}\pars{x} + 25} = {\pi \over 2} + 15\,{1 \over 25}\,{5 \over 3}\int_{0}^{\pi/2}{\bracks{3\sec^{2}\pars{x}/5}\,\dd x \over \bracks{3\tan\pars{x}/5}^{\, 2} + 1} \\[5mm] = &\ {\pi \over 2}\ +\ \underbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}} _{\ds{=\ {\pi \over 2}}}\ =\ \bbx{\pi} \end{align}

Answer 3

Sugerencia: Sustituir $$\cos(x)=\frac{1-t^2}{1+t^2}$$ y $$dx=\frac{2}{1+t^2}dt$$