Evaluate: ∫3π/4π/4 dx/(1+cosx)

Let I = ∫3π/4π/4 dx/(1+cosx)
I = ∫3π/4π/4 dx/(1+2cos2(x/2)-1)
I = ∫3π/4π/4 dx/(2cos2(x/2))
I = 1/2 ∫3π/4π/4 sec2(x/2) dx
I = 1/2 × 1/2 [tan(x/2)]3π/4π/4
I = [tan(3π/8) - tan(π/8)]
I = √2+1 - (√2-1)
I = 2
Hence, ∫3π/4π/4 dx/(1+cosx) = 2