integration?

∫ (1-cos x) / (1+cos x) dx from x=0 to π/2

= ∫ ( (1-cos x) / 2 ) / ( (1+cos x) / 2) dx from x=0 to π/2

= ∫ (sin^2 x/2) / (cos^2 x/2) dx from x=0 to π/2

= ∫ tan^2 x/2 dx from x=0 to π/2

= ∫ ( sec^2 x/2 - 1 ) dx from x=0 to π/2

= [ 2 tan x/2 - x ] from x=0 to π/2

= 2 - π/2

Mind (1-cos x)/(1+cos x)=((1-cos x)/2)/((1+cos x)/2)=(sin^2 x/2)/(cos^2 x/2)=tan^2 x/2=sec^2 x/2 - 1.
Hence ∫ (1-cos x)/(1+cos x) = ∫ (sec^2 x/2 - 1) dx = 2 tan x/2 - x + C.

And then plug in 0, pi/2, you can get:
(2 - pi/2) - 0 = 2 - pi/2.