A calculus problem by Lorenzo Di Patrizi

$\begin{aligned} I & = \int_0^\pi \frac x{1+\sin x} dx & \small \color{#3D99F6}{\text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^\pi \frac x{1+\sin x} + \frac {\pi -x}{1+\color{#3D99F6}{\sin (\pi -x)}} dx & \small \color{#3D99F6}{\text{Note that }\sin (\pi-x) = \sin x} \\ & = \frac 12 \int_0^\pi \frac x{1+\sin x} + \frac {\pi -x}{1+\color{#3D99F6}{\sin x}} dx \\ & = \frac 12 \int_0^\pi \frac \pi{1+\color{#3D99F6}{\sin x}} dx & \small \color{#3D99F6}{\text{Let }t = \tan \frac x2, \ \sin x = \frac {2t}{1+t^2}, \ dx = \frac {2\ dt}{1+t^2}} \\ & = \frac \pi 2 \int_0^\infty \frac 2{(1+t^2)\left(1+\frac {2t}{1+t^2} \right)} dt \\ & = \pi \int_0^\infty \frac 1{(1+t)^2} dt \\ & = \frac \pi {1+t} \bigg|_\infty^0 = \boxed{\pi} \approx \boxed{3.14} \end{aligned}$