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

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

Answer 1

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.

Answer 2

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: $$ I = \int_0^{\pi/2} \frac {\mathrm dx} {1+\tan(x)^\pi} = \int_0^{\pi/2} \frac {\mathrm dx} {1 + \cot(x)^\pi} = \int_0^{\pi/2} \frac {\tan(x)^\pi \mathrm dx}{1+ \tan(x)^\pi} \implies 2I = \frac \pi 2 \implies I = \frac \pi 4. $$

Answer 3

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 .

Answer 4

The usual method for integrals of this form is to use

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