$\int\limits_0^{\pi/2n}\frac{dx}{1+(\tan nx)^n}$ is equal to n ∈ N.

$\frac{n\pi}{4}$ $\frac{\pi}{2n}$ $\frac{\pi}{4n}$ $\frac{2\pi}{n}$

$\frac{\pi}{4n}$

$I=\int\limits_0^{\pi/2n}\frac{dx}{1+(\tan nx)^n}=\int\limits_0^{\pi/2n}\frac{dx}{1+(\tan n(\frac{\pi}{2n}-x))^n}=\int\limits_0^{\pi/2n}\frac{(\tan nx)^n}{1+(\tan nx)^n}⇒2I=\int\limits_0^{\pi/2n}dx⇒I=\frac{\pi}{4n}$