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This is the main problem

$$\int_{0}^{\pi} \frac{\sin{x}}{1+\cos^{2}{x}}\,dx$$

and if I let $u=\cos{x}$ then

$$-\int_{1}^{-1}\frac{du}{1+u^{2}}$$

which equals $-\tan^{-1}{u}$ from $1$ to $-1$

and where I make the mistake that I don't understand is

$$-\tan^{-1}(-1) = -\frac{3\pi}{4}$$

but this gives me the wrong answer, In order to get the correct one I'm supposed to get that

$$-\tan^{-1}(-1) = \frac{\pi}{4}$$

This has been bugging me for quite some time. is there some weird rule that says I have to use $-\frac{\pi}{4}$ instead of $\frac{3\pi}{4}$

egreg
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Nik1645
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  • Yes, there is a weird rule. It says that arctangent is the principal branch, between $-\pi/2$ and $\pi/2$ This results in an arctan function that is continuous. – Will Jagy Apr 08 '16 at 22:04
  • @WillJagy I don't think it is that weird. It's the same principle which rules for other trigonometric functions: restrict their domains as to make them injective and thus invertible. – DonAntonio Apr 08 '16 at 22:06
  • @Joanpemo He is referring to the asker's description of this rule as "weird". – Edward Evans Apr 08 '16 at 22:07
  • @Joanpemo right, the person asking the question wondered about "is there some weird rule" – Will Jagy Apr 08 '16 at 22:07
  • @WillJagy Thank you. I thought you were agreeing with the asker about the "weird rule" thing...today I'm a little sarcasm-catching impaired – DonAntonio Apr 08 '16 at 22:08
  • @Joanpemo Got it. I have slowed down on the sarcasm over the years on this site, it is often too much effort. This time I was given the language to use and did not need to think about it. – Will Jagy Apr 08 '16 at 22:12

3 Answers3

4

There are multiple branches for $\tan^{-1}(x)$ each differing from another by an integer multiple of $\pi$.

The principal branch, which is continuous on the interval from $-1$ to $1$, gives $\tan^{-1}(-1)=-\frac\pi4$ and $\tan^{-1}(1)=\frac\pi4$. Thus, the integral should be $$ \tan^{-1}(1)-\tan^{-1}(-1)=\frac\pi2 $$

The problem with using $\tan^{-1}(-1)=\frac{3\pi}4$ is that that is from a different branch than the one for which $\tan^{-1}(1)=\frac\pi4$. If you want to use $\tan^{-1}(-1)=\frac{3\pi}4$, you would need to use $\tan^{-1}(1)=\frac{5\pi}4$

robjohn
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1

With tangent we have to restrict its domain so as to be sure it is injective, and it is usually chosen the domain $\;\left(-\frac\pi2,\,\frac\pi2\right)\;$, which is then the codomain of its inverse function $\;\arctan\;$ .

As for the integral: observe that $\;\sin x=-(\cos x)'\;$ , so

$$\int_0^\pi\frac{\sin x\;dx}{1+\cos^2x}=\left.-\int_0^\pi\frac{d(\cos x)}{1+\cos^2x}=-\arctan\cos x\right|_0^\pi=$$

$$=-\arctan\cos\pi+\arctan\cos0=-\arctan(-1)+\arctan1=-\left(-\frac\pi4\right)+\frac\pi4=\frac\pi2$$

DonAntonio
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1

To supplement RobJon's answer, the multiple branches for the arctangent function have integral representation

$$\arctan(x;n)=n\pi+\int_0^x \frac{1}{1+u^2}\,du \tag 1$$

for integer $n$. On the principal branch for the arctangent, $n=0$.

Using $(1)$, we can write the integral of interest as

$$\int_{-1}^1\frac{1}{1+u^2}\,du=\arctan(1;n)-\arctan(-1;n) \tag 2$$

The choice of branch for the arctangent does not impact the value of the integral in $(2)$, provided one uses only one branch for both $\arctan(1;n)$ and $\arctan(-1;n)$.

An as aside, we can evaluate the integral in $(2)$ without appealing to the arctangent function. For $|x|\le 1$, the integral in $(1)$ can be transformed into the series

$$\int_0^x \frac{1}{1+u^2}\,du=\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n+1}}{2n+1}$$

For $x= 1$, Leibniz used a purely geometric proof to show

$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}=\frac{\pi}{4} $$

from which it trivially follows that

$$\int_{-1}^{1}\frac{1}{1+u^2}\,du=\frac{\pi}{2}$$

Using Leibniz's result, we can also evaluate the integral $\int_0^\infty \frac{1}{1+u^2}\,du$ by writing

$$\begin{align} \int_0^\infty \frac{1}{1+u^2}\,du&=\int_0^1\frac{1}{1+u^2}\,du+\int_1^\infty \frac{1}{1+u^2}\,du \tag 3\\\\ &=2\int_0^1 \frac{1}{1+u^2}\,du \tag 4\\\\ &=\frac{\pi}{2} \end{align}$$

where in going from $(3)$ to $(4)$ we enforced the substitution $u\to 1/u$ in the second integral. Therefore, we find that

$$\lim_{x\to \infty}\arctan(x;n)=n\pi+\pi/2$$

as expected!

Mark Viola
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