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Let's consider a problem, which is to find the indefinite integral $$I(x):=\displaystyle \int \frac{1}{2+\cos 2x}dx.$$ Since the integrand $f(x):=\dfrac{1}{2+\cos 2x}$ is continuous over $(-\infty,+\infty)$, hence $I(x)$ exists necessarily, accoring to the following theorem:

every function continuous over the given interval $I$ has its anti-derivative,namely indefinite integral over $I$.

Ok, now let's find it. The steps may be not so complicated. \begin{align*} I(x)&=\int \frac{1}{2+\cos 2x}dx\\ &=\frac{1}{2}\int\frac{1}{2+\cos 2x}d(2x)~~~&\textit{$2x=u$}\\ &=\frac{1}{2}\int\frac{1}{2+\cos u}d(u)\\ &=\frac{1}{2}\int\frac{\sec^2(u/2)}{\tan^2(u/2)+3}du~~~&\textit{$(\tan(u/2))/\sqrt{3}=v$}\\ &=\frac{1}{\sqrt{3}}\int\frac{1}{v^2+1}dv\\ &=\frac{1}{\sqrt{3}}\arctan v+C\\ &=\frac{1}{\sqrt{3}}\arctan \frac{\tan x}{\sqrt{3}}+C. \end{align*} The result comes out! Is it true? The WA outputs the same result! The problem seems to be done like this. But wait,wait,please scrutinize the form of the obtained indefinite integral. It is not continuous at every $x=2k\pi\pm \dfrac{\pi}{2}(k=0,\pm 1,\pm 2,\cdots).$ But, we know that

Every indefinite integral is necessarily continuous at the corresponding inteval.

What's $\displaystyle \int \frac{1}{2+\cos 2x}dx$ on earth?

StubbornAtom
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mengdie1982
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    “it has infinity discontinuity at those points” – no, it has left and right limits $\pm \frac{2}{\pi \sqrt 3}$ at those points. – Martin R Jun 02 '19 at 07:26
  • @MartinR yes, you're right! but the problem is still there. – mengdie1982 Jun 02 '19 at 07:30
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    What interval are you talking about? Remember that when using integration by substitution one has to be careful with the integration bounds. How are you addressing that in your second substitution? – Fimpellizzeri Jun 02 '19 at 07:31
  • To make it continuous you can add a function that is piece-wise constant to it, such as $\frac 1 {\sqrt 3}\left(x+\arctan\tan(-x)\right)$ –  Jun 02 '19 at 07:32
  • @Fimpellizieri I want to find the integral over the whole interval $(-\infty,+\infty)$ – mengdie1982 Jun 02 '19 at 07:36
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    That's not a good idea. Your integrand is at least $1/3$, so such an integral does not converge. – Fimpellizzeri Jun 02 '19 at 07:39
  • @Fimpellizieri not definite integral,just indefinite integral... – mengdie1982 Jun 02 '19 at 07:40
  • Please consider what conditions an integral and substitution must satisfy in order to apply integration by substitution. – Fimpellizzeri Jun 02 '19 at 07:44
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    You have found antiderivatives on each interval $(k\pi - \frac \pi 2, k \pi +\frac\pi 2)$, which can be continuously extended to the closed intervals. By adding suitable integration constants you can stitch these together to an antiderivative on $\Bbb R$. (Nobody says that an antiderivative must have a “closed” or “nice” formula). – Martin R Jun 02 '19 at 07:47
  • A similar question: https://math.stackexchange.com/questions/1356523/what-are-the-restrictions-on-using-substitution-in-integration. I can repeat my recommendation made there, to search for the articles by David J. Jeffreys on this topic. – Hans Lundmark Jun 02 '19 at 09:49

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The primitive $F$ of $\frac1{2+\cos x}$ such that $F(0)=0$ is the map$$x\mapsto\begin{cases}\frac1{\sqrt3}\arctan\left(\frac{\tan x}{\sqrt3}\right)&\text{ if }x\in\left[0,\frac\pi2\right)\\\frac\pi{2\sqrt3}&\text{ if }x=\frac\pi2\\\frac1{\sqrt3}\arctan\left(\frac{\tan x}{\sqrt3}\right)+\frac\pi{\sqrt3}&\text{ if }x\in\left(\frac\pi2,\frac{3\pi}2\right)\\\frac{3\pi}{2\sqrt3}&\text{ if }x=\frac{3\pi}2\\\frac1{\sqrt3}\arctan\left(\frac{\tan x}{\sqrt3}\right)+\frac{2\pi}{\sqrt3}&\text{ if }x\in\left(\frac{3\pi}2,\frac{5\pi}2\right)\\\vdots\end{cases}$$and, of course, if $x<0$, then $F(x)=-F(-x)$.

But you don't need all of this if all you want is to compute $\int_{-\infty}^\infty\frac{\mathrm dx}{2+\cos x}$. Since $(\forall x\in\mathbb R):f(x)\geqslant\frac13$, that integral is equal to $\infty$.

José Carlos Santos
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  • how would you prove that $F'(x) = f(x),\forall x\in\mathbb{R}$? I know one way is to use the definition of a derivative when $x= \dfrac{(2k+1)\pi}{2},k\in\mathbb{Z},$ but that appears wayyy too tedious. I was wondering if there was another way? Or maybe another function that is much easier to differentiate? –  Feb 05 '20 at 02:11
  • Let $\varphi(x)=\int_0^x\frac{\mathrm dt}{2+\cos t}$. Then $\varphi$ is a primitive of $\frac1{2+\cos x}$ and $\varphi(0)=0$. Since, on $\left(-\frac\pi2,\frac\pi2\right)$, $F$ is also a primitive of $\frac1{2+\cos x}$ and since $F(0)=0$, $\varphi=F$ on $\left(-\frac\pi2,\frac\pi2\right)$. But then$$\lim_{x\to\frac\pi2^-}\varphi(x)=\lim_{x\to\frac\pi2^-}F(x)=\frac\pi{2\sqrt3}.$$Applying the same argument on $\left(\frac\pi2,\frac{3\pi}2\right)$, you get that $F$ and $\varphi$ are equal there and that $\varphi\left(\frac{3\pi}2\right)=\frac{3\pi}{2\sqrt3}$. And so on. – José Carlos Santos Feb 05 '20 at 09:02