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

Question

I am trying to solve the integral: $$I=\int_0^{\pi/2}\frac{1}{1+\tan^{n}x}dx$$ I have tried several methods shown below: $$I(n)=\int_0^{\pi/2}\frac{1}{1+\tan^nx}dx$$ $x=\arctan(u)$ $$I(n)=\int_0^\infty\frac{1}{1+u^n}\frac{1}{1+u^2}du$$ but this does not seem to lead anywhere. I also tried: $$I(a)=\int_0^{\pi/2}\frac{1}{1+\tan^ax}dx$$ $$I'(a)=\int_0^{\pi/2}\frac{\ln(\tan x)}{\left(1+\tan^ax\right)^2}dx$$ but this just seems to complicate it more.

I also see that it can be expressed as: $$I(b)=\int_0^{\pi/2}\frac{1}{1+\tan^b(x)}dx=\int_0^{\pi/2}\frac{\cos^b(x)}{\cos^b(x)+\sin^b(x)}dx$$ a final thought is using the identity: $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ so: $$\int_0^{\pi/2}\frac{\cos^b(x)}{\cos^b(x)+\sin^b(x)}dx=\int_0^{\pi/2}\frac{\sin^b(x)}{\sin^b(x)+\cos^b(x)}$$ and therefore: $$2I(b)=\int_0^{\pi/2}1dx=\pi/2$$ $$I(b)=\pi/4$$ $$I=\pi/4$$ does this work? Thanks

Your last attempt does work. It is $\pi/2$. I handled a very, very similar integrand here. — Parcly Taxel, Oct 07 '18 at 15:27
i agree, there would only be ambiguity if it said $\tan x^\pi$ — Henry Lee, Oct 07 '18 at 15:37
How is this a duplicate they are literally different questions — Henry Lee, Oct 07 '18 at 15:58
@HenryLee In fact, the two questions are almost identical, and the highest rated answer to the older question completely answers your question. — Xander Henderson, Oct 07 '18 at 16:28
Doesn’t that make it a useful source rather than a duplicate then? — Henry Lee, Oct 07 '18 at 16:31
The last part of this question is a duplicate. That doesn't make it a duplicate question. — TonyK, Oct 07 '18 at 19:36
I've nominated to reopen this because I challenge the duplicate. The two questions are amenable to the same solution technique, but the problems themselves are very different. Besides which, good luck finding the unique oldest problem on here solvable that way. — J.G., Oct 07 '18 at 20:29

score 2 · Answer 1 · answered Oct 07 '18 at 15:30

Seems fine, but you got a typo at the last part. It should be $I(b) = \pi/4$ not $\pi/2$. Also we could do this quicker: $$ I = \int_0^{\pi/2} \frac {\mathrm dx} {1+\tan(x)^\pi} = \int_0^{\pi/2} \frac {\mathrm dx} {1 + \cot(x)^\pi} = \int_0^{\pi/2} \frac {\tan(x)^\pi \mathrm dx}{1+ \tan(x)^\pi} \implies 2I = \frac \pi 2 \implies I = \frac \pi 4. $$

score 2 · Accepted Answer · answered Oct 12 '18 at 04:32

I would like to remark that your substitution $x = \arctan{u}$ works too; as follows:

Let $u \mapsto {u}^{-1}$ then $\displaystyle I(n)=\int_0^\infty\frac{1}{1+{u}^{-n}}\frac{1}{1+u^{-2}} \cdot \frac{1}{u^2}\,du =\int_0^\infty\frac{1}{1+{u}^{-n}}\frac{1}{1+u^{2}} \, du$

Hence $\displaystyle 2I(n) = \int_0^{\infty} \frac{1}{u^2+1}\bigg(\frac{1}{1+u^n}+\frac{1}{1+u^{-n}}\bigg)\,du$ but $\displaystyle \frac{1}{1+u^n}+\frac{1}{1+u^{-n}} = 1$.

Hence $\displaystyle 2I(n) = \int_0^{\infty} \frac{1}{u^2+1}\,{du} = \frac{\pi}{2}$ therefore $\displaystyle I(n) = \frac{\pi}{4}$ as you have correctly found.

Archis Welankar · Answer 3 · 2018-10-12T11:07:50.553

1

Use $$\int _a^b f(a+b-x)dx=\int_a^b f(x)dx$$ and see the magic happen. The answer is $\frac{\pi}{4 }$ actually it's true for any non-negative number .

edited Oct 12 '18 at 11:07

answered Oct 07 '18 at 15:29

Archis Welankar

15,910

have you read the end of my answer – Henry Lee Oct 07 '18 at 15:36
I just mean the answer in general, and why is it not true for all positive numbers? – Henry Lee Oct 07 '18 at 15:38

score 0 · Answer 4 · answered Oct 07 '18 at 15:39

0

The usual method for integrals of this form is to use

$$t=\tan\frac{u}{2}$$

answered Oct 07 '18 at 15:39

Matthew Hunt

543

that would give a polynomial to the power of and unknown number (or $\pi$) and it doesn't really work – Henry Lee Oct 07 '18 at 15:41

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

4 Answers4