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I have tried to solve this which goes as follows-
$$\begin{align*} \int\frac{\cos x+2}{5+4\cos x} dx &=\int\frac{(1/4)(5+4\cos x)+(3/4)}{5+4\cos x} dx\\ &={1\over4}\int dx+{3\over4}\int\frac{dx}{9\cos^2 {x\over 2}+\sin^2{x\over2}}\\ &=({x\over4}+C)+{3\over4}\int\frac{\sec^2{x\over2}}{9+\tan^2{x\over2}}dx\\ &=({x\over4}+C)+{3\over2}\int\frac{dy}{9+y^2}\qquad \text{substituting $\tan {x\over2}=y$}\\ &={x\over4}+{1\over2}\arctan({y\over3})+C'={x\over4}+{1\over2}\arctan({\tan{x\over2}\over3})+C' \end{align*}$$ So, $$\int_{0}^{2\pi}\frac{\cos x+2}{5+4\cos x} dx=\left[{x\over4}\right]_{0}^{2\pi}+{1\over2}\left[\arctan({\tan{x\over2}\over3})\right]_{0}^{2\pi} \\={\pi\over2}+{1\over2}\left[\arctan({\tan\pi\over3})-0\right]={\pi\over2}+{1\over2}\left[0-0\right]={\pi\over2}$$ So, I can't get $\pi$!! Can anybody find out where is my fault? Thanks for assistance in advance.

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3 Answers 3

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Note that $$\int\frac{\cos x+2}{5+4\cos x} dx=F(x):={x\over4}+{1\over2}\arctan\left({\tan{x\over2}\over3}\right)+c$$ but the function $F$ is not continuous on the interval of integration $[0,2\pi]$ (on the other hand a primitive of a continuous function should be continuous by the Fundamental Theorem of Calculus).

Since $F$ has a "jump" at $x=\pi$ we may split the interval and therefore we obtain $$\int_0^{2\pi}\frac{\cos x+2}{5+4\cos x} dx =\int_0^{\pi} +\int_{\pi}^{2\pi} =[F(x)]_{0^+}^{\pi^-}+[F(x)]_{\pi^+}^{2\pi^-}=\frac{\pi}{2}+\frac{\pi}{2}=\pi.$$

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  • $\begingroup$ @BiswarupSaha See my edit! Is it clear now? $\endgroup$
    – Robert Z
    Feb 12, 2019 at 15:11
  • $\begingroup$ But $\left[\arctan({\tan{x\over2}\over3})\right]_{0}^{\pi}={\pi\over2}$ and $\left[\arctan({\tan{x\over2}\over3})\right]_{\pi}^{2\pi}=-{\pi\over2}$? $\endgroup$
    – MathBS
    Feb 12, 2019 at 15:13
  • $\begingroup$ @BiswarupSaha You shoul take right limit at $\pi$ and left limit at $2\pi$!! $\endgroup$
    – Robert Z
    Feb 12, 2019 at 15:17
  • $\begingroup$ @greedoid $\frac{\cos x+2}{5+4\cos x}$ does exist and it is continuous on $[0,2\pi]$ $\endgroup$
    – Robert Z
    Feb 12, 2019 at 15:18
  • $\begingroup$ Ah, yes, I don't know why I saw -4/5 instead of -5/4 $\endgroup$
    – nonuser
    Feb 12, 2019 at 15:20
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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Hint: Substitute $$\cos(x)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2}{1+t^2}dt$$

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    $\begingroup$ I think the OP is asking specifically where he's made a mistake as opposed to how to integrate the function. $\endgroup$
    – Clayton
    Feb 12, 2019 at 15:00
  • $\begingroup$ “Can anybody find out where is my fault?” – once more I wonder if you read all of the question ... $\endgroup$
    – Martin R
    Feb 12, 2019 at 16:06

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