A calculus problem by Jose Sacramento

Let $\displaystyle I=\int_0^{2 \pi} \frac 1 {1+\tan ^4 x} dx$ .

We can rewrite $I$ as $\displaystyle I=\int_{-\pi}^{\pi} \frac 1 {1+\tan^4 x} dx$ .

We note that the intrgrant is an even function, therefore: $\displaystyle I=2 \int_0^{\pi} \frac 1 {1+\tan^4 x} dx$

We also note that the integrant has a reflection symmetry across $x=\frac \pi 2$ , therefore:

$\begin{aligned} I&=4\int_0^{\frac \pi 2} \frac 1 {1+\tan^4 x} dx \end{aligned}$

Using the "integration backward": $\displaystyle \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx$ , we have:

$\begin{aligned} I & =2 \int_0^{\frac \pi 2} \frac 1 {1+\tan^4 x} + \frac 1 {1 +\tan^4 \left(\frac \pi 2 - x\right)} dx \\ & =2 \int_0^{\frac \pi 2} \frac 1 {1+\tan^4 x} + \frac 1 {1 +\cot^4 x} dx \\ & =2 \int_0^{\frac \pi 2} \frac 1 {1+\tan^4 x} + \frac 1 {1 +\frac 1 {\tan^4 x}} dx \\ & =2 \int_0^{\frac \pi 2} \frac 1 {1+\tan^4 x} + \frac {\tan^4 x} {\tan^4 x+1} dx \\ & =2 \int_0^{\frac \pi 2} 1 \ dx \\ & = \pi \approx \boxed{3.141} \end{aligned}$ .