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I have tried a lot of trig substitutions for this question and it's supposed to equal $\frac{\pi}{4}$ using King's method. Could someone give me some pointers on how to go about this?

$$\int_0^{\pi/2} \frac{1}{1+(\tan{x})^{\pi}}dx$$

sophie evans
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  • Recall whet is the basis of King's mathod and you will get the result immediately. You can change $\pi$ by whatever you want. – Claude Leibovici Jun 26 '24 at 10:07
  • @ClaudeLeibovici i only learned of kings method yesterday and am a first year uni student. i have tried to sub in different trig identities after applying kings method and nothing has worked yet. coudl you possibly expand a little more on what to do/ – sophie evans Jun 26 '24 at 10:12
  • Please do some personal work before, such as including in your post your attempt(s) and/or what King's method is. – Anne Bauval Jun 26 '24 at 10:24
  • Besides lacking "context", this post is a megadupe. To avoid redundancy on this site, imo it should not be answered. – Anne Bauval Jun 27 '24 at 12:04
  • @AnneBauval well its not my problem you do not know what it is? people have answered my question just fine and now you are closing it?? – sophie evans Jun 27 '24 at 16:58
  • This kind of situation is a plague for the site. There were already plenty of answers to your question in the many dupes. Your post was not closed as a duplicate, though, but for lack of "context" (see the closure banner above). – Anne Bauval Jun 28 '24 at 05:23
  • @AnneBauval i dont think it needs too much context. its an integral i need help with. im confused on what to write and i already wrote about how i tried to solve it before. if you dont know kings method thats not my issue, dont answer it then. – sophie evans Jun 28 '24 at 08:50
  • I know what King's method is, but I did not answer your request for help because, again, answering such PSQs encourages low-quality questions on this site. It was your issue to put "context" (in your post, not just in comments). See the details and links about this duty of yours, in the closure banner above and here: "Too many questions begin or end with "I don't even know how to begin with this problem". While this may be true[...]" – Anne Bauval Jun 28 '24 at 13:58

1 Answers1

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King's property of integration says that $$\int_a^b f(x)\,dx=\int_a^bf(a+b-x)\,dx$$ We can apply this property to the given problem.

Let $$I=\int_0^{\pi/2}\frac{1}{1+(\tan x)^\pi}\,dx$$ Then, by King's property of integration, we can also say \begin{align*} I&=\int_0^{\pi/2}\frac{1}{1+(\tan(0+\frac{\pi}{2}-x))^\pi}\,dx\\ &=\int_0^{\pi/2}\frac{1}{1+(\tan(\frac{\pi}{2}-x))^\pi}\,dx\\ &=\int_0^{\pi/2}\frac{1}{1+(\cot x)^\pi}\,dx\tag{since $\tan(\pi/2-x)=\cot x$} \end{align*} Now, since the two integrals are equal, let's add them together to see if we can get a simpler integration in the equation in $I$. In general, often with definite integrals it can be useful to represent an integral in multiple forms and then add or subtract them to simplify the resulting equation in $I$ and by solving for $I$ solve the initial integral. \begin{align*} 2I&=\int_0^{\pi/2}\frac{1}{1+(\tan x)^\pi}\,dx+\int_0^{\pi/2}\frac{1}{1+(\cot x)^\pi}\,dx\\ &=\int_0^{\pi/2}\frac{1}{1+(\tan x)^\pi}+\frac{1}{1+(\cot x)^\pi}\,dx\\ &=\int_0^{\pi/2}\frac{1+(\cot x)^\pi+1+(\tan x)^\pi}{(1+(\tan x)^\pi)(1+(\cot x)^\pi)}\,dx\\ &=\int_0^{\pi/2}\frac{2+(\tan x)^\pi+(\cot x)^\pi}{1+(\cot x)^\pi+(\tan x)^\pi+(\tan x)^\pi\cdot(\cot x)^\pi}\,dx\\ &=\int_0^{\pi/2}\frac{2+(\tan x)^\pi+(\cot x)^\pi}{2+(\tan x)^\pi+(\cot x)^\pi}\,dx\tag{since $(\tan x)^\pi\cdot (\cot x)^\pi=1$}\\ &=\int_0^{\pi/2}\,dx\\ &=\Big[x\Big]_0^{\pi/2}\\ &=\frac{\pi}{2} \end{align*} Because we wanted to find $I$ at the beginning, not $2I$, we divide the result by $2$. Hence, $$I=\frac{\pi}{4}$$

Hope this helps and please feel free to ask further questions, if this answer is a bit unclear! :)

Tom
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