Easy Calculus

$\begin{aligned} I & = \int_0^\frac \pi 2 \frac 1{1+\cot x} dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \left(\frac 1{1+\cot x} + \frac 1{1+\color{#3D99F6} \tan x} \right) dx & \small \color{#3D99F6} \text{Since }\cot \left(\frac \pi 2 - \theta\right) = \tan \theta \\ & = \frac 12 \int_0^\frac \pi 2 \left({\color{#3D99F6}\frac {\tan}{1+\tan x}} + \frac 1{1+\tan x} \right) dx & \small \color{#3D99F6} \text{and }\cot \theta = \frac 1{\tan \theta} \\ & = \frac 12 \int_0^\frac \pi 2 dx = \boxed {\dfrac \pi 4} \end{aligned}$

Since $0\leq x\leq \pi/2 \ , \tan(\tan^{-1}x)=x$ . Set $x=\tan^{-1} y \implies \,dx =\frac{1}{1+y^2} \,dy$ . So $\int_{0}^{\frac{1}{2}\pi} \dfrac{\tan x}{1+\tan x}\,dx= \int_{0}^{\infty}\dfrac{y}{(1+y)(1+y^2)} \,dy$ Now $\begin{aligned} \int_{0}^{\infty}\dfrac{y}{(1+y)(1+y^2)} \,dy & =\dfrac{\pi}{2}-\int_{0}^{\infty}\dfrac{1}{(1+y)(1+y^2)} \,dy\end{aligned}$ Since $\dfrac{1}{(y+1)(y^2+1)}=\dfrac{1}{2}\left(\dfrac{1}{1+y} -\dfrac{y}{y^2+1} +\dfrac{1}{y^2+1}\right)$ Thus integrating gives $\int\dfrac{2}{(y+1)(y^2+1)} = \ln(y+1) +\tan^{-1}y-\dfrac{\ln(y^2+1)}{2}+C$ And hence $\int_{0}^{\infty} \dfrac{1}{(y+1)(y^2+1)}\,dy= \dfrac{\pi}{4}$ giving us $I = \dfrac{\pi}{2}-\dfrac{\pi}{4} =\dfrac{\pi}{4}$