When problems get tough, persist on.

The integral is equivalent to inverse tan x. Evaluating inverse tan x from 0 infinity gives pi/2.

integral of 1/tan x from 0 to infinity

you forgot to mention dt .in the end .

I used contour integration for solving this . As $\displaystyle \frac{1}{t^2+1}$ is an even function . $\int_0^{\infty}\frac{1}{t^2+1}dt=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{t^2+1}dt$ We regard the integral as a closed-contour integral along the real axis, going up at the infinity on the upper half plane, coming back to the real axis at $-\infty$ . And the integrand can be written as $\frac{1}{t^2+1}=\frac{1}{(t+i)(t-i)}$ we use the general formula $\oint_{C}\frac{f(z)}{z-z_0}dz =2\pi if(z_0)$ the above described contour encircles the pole at $i$ , and $f(z)$ in above formula can be taken as $\displaystyle \frac{1}{t+i}$ $\oint_C\frac{1}{t^2+1}dt=\oint_C\frac{\frac{1}{t+i}}{t-i}dt=2\pi i \frac{1}{i+i}=\pi$ which gives $\int_0^{\infty}\frac{1}{t^2+1}dt=\frac{\pi}{2}$

You have forgot to write dt in your integral kindly edit your question