Definite Integral

zero

Please Please Please Please.Show me the working.

I asked for working not the direct answer.

And yes ,the answer is correct.It is indeed pi/2.

\int_0^{\pi} \frac{dx}{1+\tan x} = \frac{1}{2} \int_0^{\pi} 1 + \frac{1-\tan x}{1+ \tan x} \ dx= \frac{\pi}{2} + \frac{1}{2} \int_0^{\pi} \tan \left(\frac{\pi}{4}-x \right) \ dx

\int_0^{\pi} \tan \left(\frac{\pi}{4}-x \right) \ dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \tan y \ dy = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot y \ dy = 0

Hence the integral evaluates to

\frac{\pi}{2}

but tan x is discontinous at π/2,we have to break the limits in the last step